Magnus Ehingers undervisning

— Allt du behöver för A i Biologi, Kemi, Bioteknik, Gymnasiearbete med mera.

Kompletta lösningar (med videor) till utvalda övningsuppgifter i kapitel F i Ehinger: Katalys kemi 1 (Gleerups).

Hoppa direkt till uppgift: F5, F9, F11, F16, F17, F18, F21, F25, F26, F27, F29, F30F31F32F33F34F35F36, F37F40, F42F43F44F45F46F47, F48, F49, F50F51F52F53F54F55F56, F59, F60

Kapitel F

Vad gäller om gällande siffror?

Gällande siffror

Antalet gällande siffror beror på mätningens precision och noggrannhet.

Exempel på olika antal gällande siffror

123 g

Tre gällande siffror

1002 cm3

Fyra gällande siffror

12,00 dm3

Fyra gällande siffror (även nollorna efter decimalkommat är gällande, eftersom de anger hur noga mätningen skett)

0,0012 mol

Två gällande siffror (de inledande nollorna berättar bara om storleksordningen på talet, inte hur noga mätningen skett)

1200 ml

Huvudsakligen fyra gällande siffror (ibland även tre eller två beroende på sammanhang) 

Vad gäller om avrundningar?

Dina svar ska alltid avrundas till lämpligt antal gällande siffror.

  • Behåll alla siffror i miniräknaren!
  • Räkna aldrig med avrundade siffror; ”återanvänd” aldrig avrundade svar!

Addition och subtraktion avrundas till samma antal decimaler som det tal som har lägst antal decimaler.

  • Exempel: 12,34g + 0,567g = 12,907g ≈ 12,91g

Multiplikation och division avrundas till samma antal gällande siffror som det tal som har lägst antal gällande siffror.

  • Exempel: 12,3m/s · 0,45s = 5,535m ≈ 5,5m

F5.

\[m = \rho \cdot V =11,3 \frac {\mathrm{g}}{\mathrm{cm^3}} \cdot 15,0\mathrm{cm^3} = 169,5\mathrm{g} ≈ 170\mathrm{g}\]

Svar: \(m = 170\mathrm{g}\).

F9

Formeln för kaliumjodid är KI.

\[\text{procenthalt} = \frac {m_\mathrm{1K-atom}}{m_\mathrm{1KI-enhet}} = \frac {39,10\mathrm{u}}{(39,10+126,9)\mathrm{u}} = 0,23554217 ≈ 23,55\%\]

F11

Andelen Mg i MgCl2:

\[\text{andel} = \frac {m_\mathrm{1Mg-atom}}{m_\mathrm{1MgCl_2-enhet}} = \frac {24,31\mathrm{u}}{(24,31+35,45 \cdot 2)\mathrm{u}} = 0,25533032\]

Massa Mg i 2,00 g MgCl2:

\[2,00\mathrm{g} \cdot 0,25533032 = 0,51066064\mathrm{g} ≈ 0,511\mathrm{g}\]

F16.

a.

\[N_\mathrm{H_2O} = n_\mathrm{H_2O} \cdot N_\mathrm{A} = 0,56\mathrm{mol} \cdot 6,022 \cdot 10^{23}/\mathrm{mol} = 3,37232 \cdot 10^{22} ≈ 3,4 \cdot 10^{22}\]

b.

\[N_\mathrm{O} = N_\mathrm{H_2O} = 3,4 \cdot 10^{22}\]

c.

\[N_\mathrm{H} = 2N_\mathrm{H_2O} = 2 \cdot 3,37232 \cdot 10^{22} = 6,74464 \cdot 10^{22} ≈ 6,7 \cdot 10^{22}\]

F17.

a.

\[m_\mathrm{Cu} = M_\mathrm{Cu} \cdot n_\mathrm{Cu} = 63,55\frac {\mathrm{g}}{\mathrm{mol}} \cdot 0,25\mathrm{mol} = 15,8875\mathrm{g} ≈ 16\mathrm{g}\]

Svar: \(m_\mathrm{Cu} = 16\mathrm{g}\)

b.

\[m_\mathrm{H_2O} = M_\mathrm{H_2O} \cdot n_\mathrm{H_2O} = (1,008 \cdot 2 + 16,00)\mathrm{g/mol} \cdot 1,5\mathrm{mol} = 27,024\mathrm{g} ≈ 27\mathrm{g}\]

Svar: \(m_\mathrm{H_2O} = 27\mathrm{g}\)

c.

\[m_\mathrm{HCl} = M_\mathrm{HCl} \cdot n_\mathrm{HCl} = (1,008 + 35,45)\mathrm{g/mol} \cdot 2,25\mathrm{mol} = 82,0305\mathrm{g} ≈ 82\mathrm{g}\]

Svar: \(m_\mathrm{H_2O} = 27\mathrm{g}\)

d.

\[\begin{aligned} m_\mathrm{CH_3OH} &= M_\mathrm{CH_3OH} \cdot n_\mathrm{CH_3OH} = (12,01 + 1,008 \cdot 4 + 16,00)\mathrm{g/mol} \cdot 9,27\mathrm{mol} = \\ &= 297,02934\mathrm{g} ≈ 297\mathrm{g} \end{aligned}\]

Svar: \(m_\mathrm{CH_3OH} = 297\mathrm{g}\)

e.

\[\begin{aligned}m_\mathrm{CuSO_4} &= M_\mathrm{CuSO_4} \cdot n_\mathrm{CuSO_4} = (63,55 + 32,07 + 16,00 \cdot 4)\mathrm{g/mol} \cdot 5,2\mathrm{mol} = \\ &= 830,024\mathrm{g} ≈ 0,83\mathrm{kg}\end{aligned}\]

Svar: \(m_\mathrm{CuSO_4} = 0,83\mathrm{kg}\)

F18

a

\[n_\mathrm{H_2O} = \frac {m_\mathrm{H_2O}}{M_\mathrm{H_2O}} = \frac {18\mathrm{g}}{(1,008 \cdot 2 + 16,00)\frac {\mathrm{g}}{\mathrm{mol}}} = 0,99911190\mathrm{mol} ≈ 1,0\mathrm{mol}\]

Svar: \(n_\mathrm{H_2O} = 1,0\mathrm{mol}\)

b

\[n_\mathrm{He} = \frac {m_\mathrm{He}}{M_\mathrm{He}} = \frac {20,0\mathrm{g}}{4,003\frac {\mathrm{g}}{\mathrm{mol}}} = 4,99625281\mathrm{mol} ≈ 5,00\mathrm{mol}\] 

Svar: \(n_\mathrm{He} = 5,00\mathrm{mol}\)

c

\[\begin{aligned} n_\mathrm{C_{12}H_{22}O_{11}} &= \frac {m_\mathrm{C_{12}H_{22}O_{11}}}{M_\mathrm{C_{12}H_{22}O_{11}}} = \frac {6,05\mathrm{g}}{(12,01 \cdot 12 + 1,008 \cdot 22 + 16,00 \cdot 11)\frac {\mathrm{g}}{\mathrm{mol}}} = \\ &= 0,01767476\mathrm{mol} ≈ 0,0177\mathrm{mol}\end{aligned}\]

Svar: \(n_\mathrm{C_{12}H_{22}O_{11}} = 0,0177\mathrm{mol}\)

d

\[n_\mathrm{Fe} = \frac {m_\mathrm{Fe}}{M_\mathrm{Fe}} = \frac {25,0 \cdot 10^3 \mathrm{g}}{55,85\frac {\mathrm{g}}{\mathrm{mol}}} = 447,627574\mathrm{mol} ≈ 448\mathrm{mol}\]

Svar: \(n_\mathrm{Fe} = 448\mathrm{mol}\)

e

\[n_\mathrm{Fe_3O_4} = \frac {m_\mathrm{Fe_3O_4}}{M_\mathrm{Fe_3O_4}} = \frac {2,0 \cdot 10^6\mathrm{g}}{(55,85 \cdot 3 + 16,00 \cdot 4)\frac {\mathrm{g}}{\mathrm{mol}}} = 8637,44332\mathrm{mol} ≈ 8,6 \cdot 10^3\mathrm{mol}\]

 Svar: \(n_\mathrm{Fe_3O_4} = 8,6 \cdot 10^3\mathrm{mol}\)

F21

Förbränning av etanol:

C2H5OH + 3O2 → 2CO2 + 3H2O

a

Ur reaktionsformeln kan utläsas att \(n_\mathrm{C_2H_5OH}:n_\mathrm{CO_2} = 1:2\).

Därav följer att \(n_\mathrm{CO_2} = 2n_\mathrm{C_2H_5OH} = 2 \cdot 0,5\mathrm{mol} = 1\mathrm{mol}\).

b

Ur reaktionsformeln kan utläsas att \(n_\mathrm{C_2H_5OH}:n_\mathrm{O_2} = 1:3\).

Därav följer att \(n_\mathrm{O_2} = 3n_\mathrm{C_2H_5OH} = 3 \cdot 1,25\mathrm{mol} = 3,75\mathrm{mol}\).

F25

a.

  2Zn + O2 → 2ZnO
\[m\]  ①\[12\mathrm{g}\]   ④\[15\mathrm{g}\]
\[n\] ②\[0,18354237\mathrm{mol}\]   ③\[0,18354237\mathrm{mol}\]

① \(m_\mathrm{Zn} = 12\mathrm{g}\)

② \(n_\mathrm{Zn} = \frac {m_\mathrm{Zn}}{M_\mathrm{Zn}} = \frac {12\mathrm{g}}{65,38\frac {\mathrm{g}}{\mathrm{mol}}} = 0,18354237\mathrm{mol}\)

③ 

\(n_\mathrm{Zn}:n_\mathrm{ZnO} = 1:1 \\ n_\mathrm{ZnO} = n_\mathrm{Zn} = 0,18354237\mathrm{mol}\)

④ \[\begin{aligned} m_\mathrm{ZnO} &= M_\mathrm{ZnO} \cdot n_\mathrm{ZnO} = (65,38 + 16,00)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 0,18354237\mathrm{mol} = \\ &= 14,9366779\mathrm{g} ≈ 15\mathrm{g}\end{aligned}\]

b.

  2Zn + O2 → 2ZnO
\[m\]  ①\[6,5\mathrm{g}\] ④\[1,6\mathrm{g}\]  
\[n\] ②\[0,09941878\mathrm{mol}\] ③\[0,04970939\mathrm{mol}\]  

① \(m_\mathrm{Zn} = 6,5\mathrm{g}\)

② \(n_\mathrm{Zn} = \frac {6,5\mathrm{g}}{65,38\frac {\mathrm{g}}{\mathrm{mol}}} = 0,09941878\mathrm{mol}\)

③ 

\(n_\mathrm{Zn}:n_\mathrm{O_2} = 2:1 \\ n_\mathrm{O_2} = \frac {1}{2}n_\mathrm{Zn} = \frac {1}{2} \cdot 0,09941878\mathrm{mol} = 0,04970939\mathrm{mol}\)

④ \[\begin{aligned}m_\mathrm{O_2} &= M_\mathrm{O_2} \cdot n_\mathrm{O_2} = (16,00 \cdot 2)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 0,04970939\mathrm{mol} =  \\ &= 1,59070052\mathrm{g} ≈ 1,6\mathrm{g}\end{aligned}\]

F26

a.

  C2H5OH + 3O2 → 2CO2 + 3H2O
\[m\]  ①\[46\mathrm{g}\]   ④\[88\mathrm{g}\]  
\[n\] ②\[0,99852392\mathrm{mol}\]   ③\[1,99704784\mathrm{mol}\]  

① \(m_\mathrm{C_2H_5OH} = 46\mathrm{g}\)

② \(n_\mathrm{C_2H_5OH} = \frac {m_\mathrm{C_2H_5OH}}{M_\mathrm{C_2H_5OH}} = \frac {46\mathrm{g}}{(12,01 \cdot 2 + 1,008 \cdot 6 + 16,00)\frac {\mathrm{g}}{\mathrm{mol}}} = 0,99852392\mathrm{mol}\)

③ 

\(n_\mathrm{C_2H_5OH}:n_\mathrm{CO_2} = 1:2 \\ n_\mathrm{CO_2} = 2n_\mathrm{C_2H_5OH} = 2 \cdot 0,99852392\mathrm{mol} = 1,99704784\mathrm{mol}\)

④ \[\begin{aligned}  m_\mathrm{CO_2} &= M_\mathrm{CO_2} \cdot n_\mathrm{CO_2} = (12,01 + 16,00 \cdot 2)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 1,99704784\mathrm{mol} = \\ &= 87,8900755\mathrm{g} ≈ 88\mathrm{g}\end{aligned}\]

b.

  C2H5OH + 3O2 → 2CO2 + 3H2O
\[m\]  ①\[22\mathrm{g}\] ④\[46\mathrm{g}\]    
\[n\] ②\[0,47755492\mathrm{mol}\] ③\[1,43266476\mathrm{mol}\]    

① \(m_\mathrm{C_2H_5OH} = 22\mathrm{g}\)

② \(n_\mathrm{C_2H_5OH} = \frac {m_\mathrm{C_2H_5OH}}{M_\mathrm{C_2H_5OH}} = \frac {22\mathrm{g}}{(12,01 \cdot 2 + 1,008 \cdot 6 + 16,00)\frac {\mathrm{g}}{\mathrm{mol}}} = 0,47755492\mathrm{mol}\)

③ 

\(n_\mathrm{C_2H_5OH}:n_\mathrm{O_2} = 1:3 \\ n_\mathrm{O_2} = 3n_\mathrm{C_2H_5OH} = 3 \cdot 0,47755492\mathrm{mol} = 1,43266476\mathrm{mol}\)

④ \[\begin{aligned}m_\mathrm{O_2} &= M_\mathrm{O_2} \cdot n_\mathrm{O_2} = (16,00 \cdot 2)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 1,43266476\mathrm{mol} = \\ &= 45,8452722\mathrm{g} ≈ 46\mathrm{g}\end{aligned}\]

c.

  C2H5OH + 3O2 → 2CO2 + 3H2O
\[m\]    ①\[87\mathrm{g}\]  ④\[80\mathrm{g}\]  
\[n\]   ②\[2,71875\mathrm{mol}\]  ③\[1,8125\mathrm{mol}\]  

① \(m_\mathrm{O_2} = 87\mathrm{g}\)

② \(n_\mathrm{O_2} = \frac {m_\mathrm{O_2}}{M_\mathrm{O_2}} = \frac {87\mathrm{g}}{(16,00 \cdot 2)\frac {\mathrm{g}}{\mathrm{mol}}} = 2,71875\mathrm{mol}\)

③ 

\(n_\mathrm{O_2}:n_\mathrm{CO_2} = 3:2 \\ n_\mathrm{CO_2} = \frac {2}{3}n_\mathrm{O_2} = \frac {2}{3} \cdot 2,71875\mathrm{mol} = 1,8125\mathrm{mol}\)

④ \[\begin{aligned}m_\mathrm{CO_2} &= M_\mathrm{CO_2} \cdot n_\mathrm{CO_2} = (12,01 + 16,00 \cdot 2)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 1,8125\mathrm{mol} \\ &= 79,768125\mathrm{g} ≈ 80\mathrm{g}\end{aligned}\]

F27

a.

  2Fe2O3 + 3C → 4Fe + 3CO2
\[m\]  ①\[25\mathrm{kg}\] ④\[2,8\mathrm{kg}\]    
\[n\] ②\[156,543519\mathrm{mol}\] ③\[234,815279\mathrm{mol}\]    

① \(m_\mathrm{Fe_2O_3} = 25\mathrm{kg} = 25 \cdot 10^3\mathrm{g} = 25000\mathrm{g}\)

② \(n_\mathrm{Fe_2O_3} = \frac {m_\mathrm{Fe_2O_3}}{M_\mathrm{Fe_2O_3}} = \frac {25000\mathrm{g}}{(55,85 \cdot 2 + 16,00 \cdot 3)\frac {\mathrm{g}}{\mathrm{mol}}} = 156,543519\mathrm{mol}\)

③ 

\(n_\mathrm{Fe_2O_3}:n_\mathrm{C} = 2:3 \\ n_\mathrm{C} = \frac {3}{2}n_\mathrm{Fe_2O_3} = \frac {3}{2} \cdot 156,543519\mathrm{mol} = 234,815279\mathrm{mol}\)

④ \(m_\mathrm{C} = 12,01\frac {\mathrm{g}}{\mathrm{mol}} \cdot 234,815279\mathrm{mol} = 2820,13150\mathrm{g} = 2,82013150 \cdot 10^3\mathrm{g} ≈ 2,8\mathrm{kg}\)

b.

  2Fe2O3 + 3C → 4Fe + 3CO2
\[m\]  ①\[500\mathrm{kg}\]   ④\[350\mathrm{kg}\]  
\[n\] ②\[3130,87038\mathrm{mol}\]   ③\[6261,74076\mathrm{mol}\]  

① \(m_\mathrm{Fe_2O_3} = 500\mathrm{kg} = 500 \cdot 10^3\mathrm{g}\)

② \(n_\mathrm{Fe_2O_3} = \frac {m_\mathrm{Fe_2O_3}}{M_\mathrm{Fe_2O_3}} = \frac {500 \cdot 10^3\mathrm{g}}{(55,85 \cdot 2 + 16,00 \cdot 3)\frac {\mathrm{g}}{\mathrm{mol}}} = 3130,87038\mathrm{mol}\)

③ 

\(n_\mathrm{Fe_2O_3}:n_\mathrm{Fe} = 1:2 \\ n_\mathrm{Fe} = 2n_\mathrm{Fe_2O_3} = 2 \cdot 3130,87038\mathrm{mol} = 6261,74076\mathrm{mol}\)

\[\begin{aligned} m_\mathrm{Fe} &= M_\mathrm{Fe} \cdot n_\mathrm{Fe} = 55,85\frac {\mathrm{g}}{\mathrm{mol}} \cdot 6261,74076\mathrm{mol} = 349718,222\mathrm{g} = \\ &= 349,718222 \cdot 10^3\mathrm{g} ≈ 350\mathrm{kg}\end{aligned}\]

c.

  2Fe2O3 + 3C → 4Fe + 3CO2
\[m\]      ①\[572\mathrm{ton}\] ④\[338\mathrm{ton}\]
\[n\]     ②\[1,02417189 \cdot 10^7\mathrm{mol}\] ③\[7,68128917 \cdot 10^6\mathrm{mol}\]

① \(m_\mathrm{Fe} = 572\mathrm{ton} = 572 \cdot 10^6\mathrm{g}\)

② \(n_\mathrm{Fe} = \frac {m_\mathrm{Fe}}{M_\mathrm{Fe}} = \frac {572 \cdot 10^6\mathrm{g}}{55,85\frac {\mathrm{g}}{\mathrm{mol}}} = 1,02417189 \cdot 10^7\mathrm{mol}\)

③ 

\(n_\mathrm{Fe}:n_\mathrm{CO_2} = 4:3 \\ n_\mathrm{CO_2} = \frac {3}{4}n_\mathrm{Fe} = \frac {3}{4} \cdot 1,02417189 \cdot 10^7\mathrm{mol} = 7,68128917 \cdot 10^6\mathrm{mol}\)

\(\begin{aligned}m_\mathrm{CO_2} &= M_\mathrm{CO_2} \cdot n_\mathrm{CO_2} = (12,01 + 16,00 \cdot 2)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 7,68128917 \cdot 10^6\mathrm{mol} = \\ &= 338,053536 \cdot 10^6\mathrm{g} ≈ 338\mathrm{ton}\end{aligned}\)

F29

a.

CoCl2 · 6H2O(s) \( \overset{\Delta}{\rightarrow}\) CoCl2(s) + 6H2O(g)

b.

\(n_\mathrm{CoCl_2 \cdot 6H_2O}:n_\mathrm{H_2O} = 1:6\) ⇒ Det bildas 6 mol.

c.

  CoCl2 · 6H2O(s) → CoCl2(s) + 6H2O(g)
\[m\]  ①\[1,0\mathrm{g}\]   ④\[0,45\mathrm{g}\]
\[n\] ②\[0,00420299\mathrm{mol}\]   ③\[0,02521792\mathrm{mol}\]

① \(m_\mathrm{CoCl_2 \cdot 6H_2O} = 1,0\mathrm{g}\)

\(\begin{aligned}n_\mathrm{CoCl_2 \cdot 6H_2O} &= \frac {m_\mathrm{CoCl_2 \cdot 6H_2O}}{M_\mathrm{CoCl_2 \cdot 6H_2O}} = \frac {1,0\mathrm{g}}{(58,93+25,45 \cdot 2 + 6 \cdot (1,008 \cdot 2 + 16,00))\mathrm{g/mol}} = \\ &= 0,00420299\mathrm{mol}\end{aligned}\)

③ 

\(n_\mathrm{CoCl_2 \cdot 6H_2O}:n_\mathrm{H_2O} = 1:6 \\ n_\mathrm{H_2O} = 6n_\mathrm{CoCl_2 \cdot 6H_2O} = 6 \cdot 0,00420299\mathrm{mol} = 0,02521792\mathrm{mol}\)

\(\begin{aligned}m_\mathrm{H_2O} &= M_\mathrm{H_2O} \cdot n_\mathrm{H_2O} = (1,008 \cdot 2 + 16,00)\mathrm{g/mol} \cdot 0,02521792\mathrm{mol} = \\ &= 0,45432614\mathrm{g} ≈ 0,45\mathrm{g}\end{aligned}\)

F30

NiSO4 · xH2O(s) \( \overset{\Delta}\rightarrow\) NiSO4(s) + xH2O(g)

\[\begin{aligned} n_\mathrm{NiSO_4} &= \frac {m_\mathrm{NiSO_4}}{M_\mathrm{NiSO_4}} = \frac {0,590\mathrm{g}}{(58,69+32,07+16,00 \cdot 4)\mathrm{g/mol}} = \\ &= 0,00381235\mathrm{mol} \\ m_\mathrm{H_2O} &= m_\mathrm{NiSO_4 \cdot xH_2O} - m_\mathrm{NiSO_4} = 1,00\mathrm{g} - 0,590\mathrm{g} = 0,410\mathrm{g} \\ n_\mathrm{H_2O} &= \frac {m_\mathrm{H_2O}}{M_\mathrm{H_2O}} = \frac {0,410\mathrm{g}}{(1,008 \cdot 2 + 16,00)\mathrm{g/mol}} =0,02275755\mathrm{mol} \\ x &= \frac {n_\mathrm{H_2O}}{n_\mathrm{NiSO_4}} = \frac {0,02275755\mathrm{mol}}{0,00381235\mathrm{mol}} = 5,96942078 ≈ 6\end{aligned}\] 

Svar: NiSO4 · 6H2O(s)

F31

FeSO4 · 7H2O(s) \(\overset{\Delta}{\rightarrow}\) FeSO4(s) + 7H2O(g)

\[\begin{aligned} n_\mathrm{FeSO_4 \cdot 7H_2O} &= \frac {m_\mathrm{FeSO_4 \cdot 7H_2O}}{M_\mathrm{FeSO_4 \cdot 7H_2O}} =  \\ &= \frac {1,26\mathrm{g}}{(55,85+32,07+16,00\cdot 4+7\cdot (1,008\cdot 2+16,00))\mathrm{g/mol}} = \\ &= 0,00453185\mathrm{mol}\end{aligned}\]

\(n_\mathrm{FeSO_4 \cdot 7H_2O}:n_\mathrm{FeSO_4} = 1:1\)
\(n_\mathrm{FeSO_4} = n_\mathrm{FeSO_4 \cdot 7H_2O} = 0,00453185\mathrm{mol}\)

\[\begin{aligned} m_\mathrm{FeSO_4} &= M_\mathrm{FeSO_4} \cdot n_\mathrm{FeSO_4} = \\ &= (55,85+32,07+16,00\cdot 4)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 0,00453185\mathrm{mol} = \\ &= 0,68847902\mathrm{g} ≈ 0,688\mathrm{g}\end{aligned}\]

F32

a.

\[n_\mathrm{Sn} = \frac {m_\mathrm{Sn}}{M_\mathrm{Sn}} = \frac {11,9\mathrm{g}}{118,7\mathrm{g/mol}} = 0,10025274\mathrm{mol} ≈ 0,100\mathrm{mol}\]

b.

\[m_\mathrm{O} = m_\mathrm{tennoxid} - m_\mathrm{Sn} = (15,1 - 11,9)\mathrm{g} = 3,2\mathrm{g}\]

c.

\[n_\mathrm{O} = \frac {m_\mathrm{O}}{M_\mathrm{O}} = \frac {3,2\mathrm{g}}{16,00\mathrm{g/mol}} = 0,20\mathrm{mol}\]

d.

\[\frac {n_\mathrm{O}}{n_\mathrm{Sn}} = \frac {0,20\mathrm{mol}}{0,100\mathrm{mol}} = 2 \\ n_\mathrm{O}:n_\mathrm{Sn} = 2:1\]

Det är dubbelt så många syreatomer som tennatomer i tennoxiden. Därför är tennoxidens formel SnO2.

F33

\(n_\mathrm{Cu} = \frac {m_\mathrm{Cu}}{M_\mathrm{Cu}} = \frac {0,252\mathrm{g}}{63,55\mathrm{g/mol}} = 0,00396538\mathrm{mol}\)

\(n_\mathrm{S} = \frac {m_\mathrm{S}}{M_\mathrm{S}}\)

\(m_\mathrm{S} = m_\mathrm{kopparsulfid} - m_\mathrm{Cu} = = 0,316\mathrm{g} - 0,252\mathrm{g} = 0,064\mathrm{g}\)

\(n_\mathrm{S} = \frac {m_\mathrm{S}}{M_\mathrm{S}} = \frac{0,064\mathrm{g}}{32,07\mathrm{g/mol}} = 0,00199563\mathrm{mol}\)

\(\frac{n_\mathrm{Cu}}{n_\mathrm{S}} = \frac{0,00396538\mathrm{mol}}{0,00199563\mathrm{mol}} = 1,98702793  ≈ \frac{2}{1}\)

\(n_\mathrm{Cu}:n_\mathrm{S} = 2:1\)

Svar: Det är dubbelt så många Cu som S i en kopparsulfidenhet. Därför blir kopparsulfidens formel Cu2S.

F34

  1. \(m_\mathrm{Pb} = 86,6\% \cdot 100\mathrm{g}\)
    \(m_\mathrm{O} = 13,4\% \cdot 100\mathrm{g}\)
  2. \(n_\mathrm{Pb} = \frac {m_\mathrm{Pb}}{M_\mathrm{Pb}} = \frac {86,6\mathrm{g}}{207,2\mathrm{g/mol}} = 0,41795367\mathrm{mol} ≈ 0,418\mathrm{mol}\)
    \(n_\mathrm{O} = \frac {m_\mathrm{O}}{M_\mathrm{O}} = \frac {13,4\mathrm{g}}{16,00\mathrm{g/mol}} = 0,8375\mathrm{mol} ≈ 0,838\mathrm{mol}\)
  3. \(\frac {n_\mathrm{Pb}}{n_\mathrm{O}} = \frac {0,41795367\mathrm{mol}}{0,8375\mathrm{mol}} ≈ \frac {0,4\mathrm{mol}}{0,8\mathrm{mol}} = \frac {1}{2}\)
    \(n_\mathrm{Pb}:n_\mathrm{O} = 1:2\)
    Svar: Blyoxidens formel är PbO2.

F35

Vi antar att vi har 100 g av ämnet.

\[\begin{aligned} m_\mathrm{C} &= 55\% \cdot 100\mathrm{g} = 55\mathrm{g} \\ n_\mathrm{C} &= \frac {m_\mathrm{C}}{M_\mathrm{C}} = \frac {55\mathrm{g}}{12,01\mathrm{g/mol}} = 4,57951707\mathrm{mol} \\ m_\mathrm{O} &= 36\% \cdot 100\mathrm{g} = 36\mathrm{g} \\ n_\mathrm{O} &= \frac {m_\mathrm{O}}{M_\mathrm{O}} = \frac {36\mathrm{g}}{16,00\mathrm{g/mol}} = 2,25\mathrm{mol} \\ m_\mathrm{H} &= 9\% \cdot 100\mathrm{g} = 9\mathrm{g} \\ n_\mathrm{H} &= \frac {m_\mathrm{H}}{M_\mathrm{H}} = \frac {9\mathrm{g}}{1,008\mathrm{g/mol}} = 8,92857143\mathrm{mol}\end{aligned}\]

\(n_\mathrm{C}:n_\mathrm{O}:n_\mathrm{H} ≈ 4,5:2,25:9 = 2:1:4\)

Svar: Ämnets empiriska formel är C2H4O (till exempel acetaldehyd, CH3CHO).

F36

\[M_\mathrm{CH} = (12,01 + 1,008)\mathrm{g/mol} = 13,018\mathrm{g/mol} \\ \frac {M_\mathrm{faktisk}}{M_\mathrm{CH}} = \frac {78,1\mathrm{g/mol}}{13,018\mathrm{g/mol}} = \frac {M_\mathrm{faktisk}}{M_\mathrm{CH}} = \frac {78,1\mathrm{g/mol}}{13,018\mathrm{g/mol}} = 5,99938547 ≈ 6\]

Ämnets molekylformel: (CH)6 C6H6

F37

a.

Vi antar att vi har 100g av ämnet.

\[\begin{aligned} m_\mathrm{C} &= 85,6\% \cdot 100\mathrm{g}=85,6\mathrm{g} \\ n_\mathrm{C} &= \frac {m_\mathrm{C}}{M_\mathrm{C}} = \frac {85,6\mathrm{g}}{12,01\frac {\mathrm{g}}{\mathrm{mol}}} = 7,12739384\mathrm{mol} \\ m_\mathrm{H} &= 14,4\% \cdot 100\mathrm{g} =14,4\mathrm{g} \\ n_\mathrm{H} &= \frac {m_\mathrm{H}}{M_\mathrm{H}} = \frac {14,4\mathrm{g}}{1,008\frac {\mathrm{g}}{\mathrm{mol}}} = 14,2857143\mathrm{mol}\end{aligned}\]

\[n_\mathrm{C}:n_\mathrm{H} ≈ 7:14 = 1:2\]

Den empiriska formeln måste vara CH2.

b.

\[\begin{aligned} M_\mathrm{faktisk} &= 70,1\mathrm{g/mol}  \\ M_\mathrm{CH_2} &= (12,01 + 1,008 \cdot 2)\mathrm{g/mol} = 14,026\mathrm{g/mol} \\ \frac {M_\mathrm{faktisk}}{M_\mathrm{CH_2}} &= \frac {70,1\mathrm{g/mol}}{14,026\mathrm{g/mol}} = 4,99786112 ≈ 5\end{aligned}\]

Molekylformeln är (CH2)5 ⇔ C5H10.

F40

a.

\[c = \frac {1\mathrm{matsked}}{3,0\mathrm{l}} = \frac {1}{3}\mathrm{matsked/l}\]

b.

\[c = \frac {m}{V} = \frac {22\mathrm{g}}{3,0\mathrm{dm^3}} = 7,33333333\mathrm{g/dm^3} ≈ 7,3\mathrm{g/dm^3}\]

c.

\[c = \frac {m}{V} = \frac {22\mathrm{g}}{3000\mathrm{ml}} = 0,00733333 ≈ 0,73\%\]

d.

\(c_\mathrm{NaCl} = \frac {n_\mathrm{NaCl}}{V}\)

\[n_\mathrm{NaCl} = \frac {m_\mathrm{NaCl}}{M_\mathrm{NaCl}} = \frac {22\mathrm{g}}{(22,99+35,45)\frac {\mathrm{g}}{\mathrm{mol}}} = 0,37645448\mathrm{mol}\]

\(c_\mathrm{NaCl} = \frac {0,37645448\mathrm{mol}}{3,0\mathrm{dm^3}} = 0,12548483\mathrm{mol/dm^3} ≈ 0,13\mathrm{M}\)

F42

a.

\[\begin{aligned}n_\mathrm{NaCl} &= \frac {m_\mathrm{NaCl}}{M_\mathrm{NaCl}} = \frac {0,50\mathrm{g}}{(22,99 + 35,45)\mathrm{g/mol}} = 0,00855578\mathrm{mol} \\ c_\mathrm{NaCl} &= \frac {n_\mathrm{NaCl}}{V} = \frac {0,00855578\mathrm{mol}}{1\mathrm{dm^3}} = 0,00855578\mathrm{mol/dm^3} ≈ 0,0086\mathrm{M}\end{aligned}\]

b.

\[\begin{aligned} n_\mathrm{C_2H_5OH} &= \frac {m_\mathrm{C_2H_5OH}}{M_\mathrm{C_2H_5OH}} = \frac {0,125\mathrm{g}}{(12,01 \cdot 2 + 1,008 \cdot 6 + 16,00)\mathrm{g/mol}} = 0,00271338\mathrm{mol} \\ c_\mathrm{C_2H_5OH} &= \frac {n_\mathrm{C_2H_5OH}}{V} = \frac {0,00271338\mathrm{mol}}{0,250\mathrm{dm^3}} = 0,01085352\mathrm{mol/dm^3} ≈ 0,0109\mathrm{M}\end{aligned}\]

c.

\[\begin{aligned} n_\mathrm{CH_3COOH} &= \frac {m_\mathrm{CH_3COOH}}{M_\mathrm{CH_3COOH}} = \frac {33\mathrm{g}}{(12,01 \cdot 2 + 1,008 \cdot 4 + 16,00 \cdot 2)\mathrm{g/mol}} =\\&= 0,00549524\mathrm{mol} \\ c_\mathrm{CH_3COOH} &= \frac {n_\mathrm{CH_3COOH}}{V} = \frac {0,00549524\mathrm{mol}}{0,650\mathrm{dm^3}} = 0,00845421\mathrm{mol/dm^3} ≈ 0,0085\mathrm{M}\end{aligned}\]

d.

\[\begin{aligned} n_\mathrm{C_6H_{12}O_6} &= \frac {m_\mathrm{C_6H_{12}O_6}}{M_\mathrm{C_6H_{12}O_6}} = \frac {0,045\mathrm{g}}{(12,01 \cdot 6 + 1,008 \cdot 12 + 16,00 \cdot 6)\mathrm{g/mol}} =\\ &= 0,00024978\mathrm{mol} \\ c_\mathrm{C_6H_{12}O_6} &= \frac {n_\mathrm{C_6H_{12}O_6}}{V} = \frac {0,00024978\mathrm{mol}}{0,016\mathrm{dm^3}} = 0,01561147\mathrm{mol/dm^3} ≈ 0,016\mathrm{M}\end{aligned}\]

F43

a.

\[\begin{aligned} c_\mathrm{CuCl_2} &= \frac {n_\mathrm{CuCl_2}}{V} \\ &n_\mathrm{CuCl_2} = \frac {m_\mathrm{CuCl_2}}{M_\mathrm{CuCl_2}}  = \frac {1,34\mathrm{g}}{(63,55 + 35,45 \cdot 2)\mathrm{g/mol}} = 0,00996653\mathrm{mol} \\ c_\mathrm{CuCl_2} &= \frac {0,00996653\mathrm{mol}}{0,100\mathrm{dm^3}} = 0,09966530\mathrm{mol/dm^3} ≈ 0,100\mathrm{mol/dm^3} \end{aligned}\]

b.

Kopparkloriden löses i vatten: CuCl2(s) Cu2+(aq) + 2Cl(aq)

\(n_\mathrm{Cu^{2+}}:n_\mathrm{CuCl_2} = 1:1 \\ [\mathrm{Cu^{2+}}] = c_\mathrm{CuCl_2} = 0,100\mathrm{mol/dm^3}\)

c.

\(n_\mathrm{Cl^-}:n_\mathrm{CuCl_2} = 2:1 \\ [\mathrm{Cl^-}] = 2c_\mathrm{CuCl_2} = 2 \cdot 0,100\mathrm{mol/dm^3} = 0,200\mathrm{mol/dm^3}\)

d. 

Eftersom all CuCl2 lösts upp, finns det inte längre kvar några CuCl2-partiklar. Därmed blir \([\mathrm{CuCl_2}] = 0\mathrm{mol/dm^3}\).

F44

a.

\[\begin{aligned} n_\mathrm{MgCl_2} &= c_\mathrm{MgCl_2} \cdot V = 0,125\frac {\mathrm{mol}}{\mathrm{dm^3}} \cdot 0,250\mathrm{dm^3} = 0,03125\mathrm{mol} ≈ 0,0313\mathrm{mol}\end{aligned}\]

b.

När magnesiumklorid löses i vatten, sker det enligt följande reaktionsformel:

MgCl2(s) Mg2+(aq) + 2Cl(aq)

\[n_\mathrm{Mg^{2+}}:n_\mathrm{MgCl_2} = 1:1 \\ [\mathrm{Mg^{2+}}] = c_\mathrm{MgCl_2} = 0,125\mathrm{mol/dm^3} \\ n_\mathrm{Mg^{2+}} = [\mathrm{Mg^{2+}}] \cdot V = 0,125\frac {\mathrm{mol}}{\mathrm{dm^3}} \cdot 0,250\mathrm{dm^3} = 0,03925\mathrm{mol} ≈ 0,0393\mathrm{mol}\]

c.

När magnesiumklorid löses i vatten, sker det enligt följande reaktionsformel:

MgCl2(s) Mg2+(aq) + 2Cl(aq)

\[n_\mathrm{MgCl_2}:n_\mathrm{Cl^-} = 1:2 \\ [\mathrm{Cl^-}] = 2c_\mathrm{MgCl_2} = 2 \cdot 0,125\mathrm{mol/dm^3} = 0,250\mathrm{mol/dm^3} \\ n_\mathrm{Cl^-} = [\mathrm{Cl^-}] \cdot V = 0,250\frac {\mathrm{mol}}{\mathrm{dm^3}} \cdot 1,150\mathrm{dm^3}  = 0,2875\mathrm{mol} ≈ 0,288\mathrm{mol}\]

d.

\[\begin{aligned} n_\mathrm{MgCl_2} &= c_\mathrm{MgCl_2} \cdot V = 0,125\frac {\mathrm{mol}}{\mathrm{dm^3}} \cdot 0,500\mathrm{dm^3} = 0,0625\mathrm{mol} \\ m_\mathrm{MgCl_2} &= M_\mathrm{MgCl_2} \cdot n_\mathrm{MgCl_2} = (24,31 + 35,45 \cdot 2)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 0,0625\mathrm{mol} = 5,950625\mathrm{g} ≈ \\ &≈ 5,95\mathrm{g} \end{aligned}\]

e.

\[\begin{aligned} c_\mathrm{MgCl_2} &= \frac {m_\mathrm{MgCl_2}}{V} = \frac {M_\mathrm{MgCl_2} \cdot n_\mathrm{MgCl_2}}{V} = \\ &= \frac {(24,31+35,45 \cdot 2)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 0,125\mathrm{mol}}{1\mathrm{dm^3}} = 11,90125\mathrm{g/dm^3} ≈ 11,9\mathrm{g/dm^3} \end{aligned}\]

F45

a.

\[V = \frac {n_\mathrm{(NH_4)_2CO_3}}{c_\mathrm{(NH_4)_2CO_3}} = \frac {1,0\mathrm{mol}}{1,56\frac {\mathrm{mol}}{\mathrm{dm^3}}} = 0,64102564\mathrm{dm^3} ≈ 0,64\mathrm{dm^3}\]

b.

Ammoniumkarbonaten löses i vatten:

(NH4)2CO3(s) 2NH\(_4^+\)(aq) + CO\(_3^{2-}\)(aq)

\(n_\mathrm{CO_3^{2-}}:n_\mathrm{(NH_4)_2CO_3} = 1:1\)

\[V = \frac {n_\mathrm{CO_3^{2-}}}{[\mathrm{CO_3^{2-}}]} = \frac {0,25\mathrm{mol}}{1,56\mathrm{mol/dm^3}} = 0,16025641\mathrm{dm^3} ≈ 0,16\mathrm{dm^3}\]

c.

Ammoniumkarbonaten löses i vatten:

(NH4)2CO3(s) 2NH\(_4^+\)(aq) + CO\(_3^{2-}\)(aq)

\(n_\mathrm{NH_4^+}:n_\mathrm{(NH_4)_2CO_3} = 2:1\)

\[\begin{aligned}\mathrm{[NH_4^+]} &= 2c_\mathrm{(NH_4)_2CO_3} = 2 \cdot 1,56 \mathrm{mol/dm^3} = 3,12\mathrm{mol/dm^3} \\ V &= \frac {n_\mathrm{NH_4^+}}{[\mathrm{NH_4^+}]} = \frac {0,75\mathrm{mol}}{3,12\mathrm{mol/dm^3}} = 0,24038462\mathrm{dm^3} ≈ 0,24\mathrm{dm^3}\end{aligned}\]

F46

\[c = \frac {m_\mathrm{paracetamol}}{m_\mathrm{tot}} = \frac {2 \cdot 500 \cdot 10^{-3}\mathrm{g}}{(93000 + 2 \cdot 500 \cdot 10^{-3}\mathrm{g})} = 1,075266 \cdot 10^{-5} ≈ 0,0011\%\]

F47

\[[\mathrm{Cl^-}] = \frac {n_\mathrm{Cl^-,tot}}{V} = \frac {n_\mathrm{Cl^-,NaCl} + n_\mathrm{Cl^-,MgCl_2}}{V}\] 

\[\begin{aligned}n_\mathrm{Cl^-,NaCl} &= n_\mathrm{NaCl} = \frac {m_\mathrm{NaCl}}{M_\mathrm{NaCl}} = \frac {1,00\mathrm{g}}{(22,99+35,45)\mathrm{g/mol}} = 0,01711157\mathrm{mol} \\ n_\mathrm{Cl^-,MgCl_2} &= 2n_\mathrm{MgCl_2} = 2\frac {m_\mathrm{MgCl_2}}{M_\mathrm{MgCl_2}} = 2 \cdot \frac {1,00\mathrm{g}}{(24,31+35,45 \cdot 2)\mathrm{g/mol}} = 0,01329610\mathrm{mol}\end{aligned}\]

\[[\mathrm{Cl^-}] = \frac {(0,01711157 + 0,01329610)\mathrm{mol}}{0,100\mathrm{dm}^3} = 0,38117764\text{mol/dm}^3 ≈ 0,381\mathrm{mol/dm^3}\]

F48

\[\begin{aligned} c = \frac {n_\mathrm{NaOH,tot}}{V_\mathrm{tot}} &= \frac {n_\mathrm{NaOH,A} + n_\mathrm{NaOH,B}}{V_\mathrm{A} + V_\mathrm{B}} = \\ &= \frac {c_\mathrm{NaOH,A} \cdot V_\mathrm{A} + c_\mathrm{NaOH,B} \cdot V_\mathrm{B}}{V_\mathrm{A} + V_\mathrm{B}} = \\ &= \frac {2,5\frac {\mathrm{mol}}{\mathrm{dm}^3} \cdot 0,030\mathrm{dm}^3 + 0,50\frac {\mathrm{mol}}{\mathrm{dm}^3} \cdot 0,075\mathrm{dm}^3}{(0,030 + 0,075)\mathrm{dm^3}} = \\ & = 1,07142857\mathrm{mol/dm^3} ≈ 1,1\mathrm{mol/dm^3} \end{aligned}\]

F49

Vi vill beräkna koncentrationen kloridjoner: 

\[[\mathrm{Cl^-}] = \frac {n_\mathrm{Cl^-}}{V}\]

Vid analysen bildas bildas fast silverklorid, AgCl(s):

Ag+(aq) + Cl(aq) AgCl(s)

Substansmängden kloridjoner beräknas:

\[\begin{aligned} n_\mathrm{Cl^-}:n_\mathrm{AgCl} &= 1:1  \\ n_\mathrm{Cl^-} &= n_\mathrm{AgCl} = \frac {m_\mathrm{AgCl}}{M_\mathrm{AgCl}} = \frac {0,253\mathrm{g}}{(107,9 + 35,45)\mathrm{g/mol}} = \\ &= 0,00176491\mathrm{mol} \end{aligned}\]

Koncentrationen kloridjoner beräknas:

\[\begin{aligned} \mathrm{[Cl^-]} &= \frac {0,00176491\mathrm{mol}}{0,250\mathrm{dm^3}} = 0,00705964\mathrm{mol/dm^3} ≈  \\ &≈ 0,00706\mathrm{mol/dm^3} = 7,06 \cdot 10^{-3}\mathrm{mol/dm^3}\end{aligned}\]

F50

\[\begin{aligned} c_1V_1 &= c_2V_2  \\ V_1 &= \frac {c_2V_2}{c_1} = \frac {2,5\mathrm{mol/dm^3} \cdot 5,0\mathrm{dm^3}}{11,8\mathrm{mol/dm^3}} = 1,05932203\mathrm{dm^3} ≈ 1,1\mathrm{dm^3}\end{aligned}\]

F51

\[\begin{aligned} c_1V_1 &= c_2V_2 \\ V_2 &= \frac {c_1V_1}{c_2} = \frac {10\% \cdot 15\mathrm{ml}}{3,0\%} = 50\mathrm{ml}\end{aligned}\]

F52

a.

\[\begin{aligned} c_1V_1 &= c_2V_2 \\ c_2 &= \frac {c_1V_1}{V_2} = \frac {0,150\mathrm{mol/dm^3}\cdot 0,025\mathrm{dm}^3}{0,350\mathrm{dm}^3} = 0,01071429 \mathrm{mol/dm}^3 ≈ \\ &≈ 0,011\mathrm{mol/dm^3}\end{aligned}\]

b.

När fast aluminiumnitrat löses i vatten, sker det enligt följande reaktionsformel:


Al(NO3)3(s) →Al3+(aq) + 3NO\(_3^-\)(aq)

\[\begin{aligned} n_\mathrm{Al(NO_3)_3}:n_\mathrm{NO_3^-} &= 1:3 \\ \mathrm{[NO_3^-]} &= 3c_\mathrm{Al(NO_3)_3}  = 3 \cdot 0,150\mathrm{mol/dm^3} = 0,450\mathrm{mol/dm^3} \\ c_1V_1 &= c_2V_2 \\ c_2 &= \frac {c_1V_1}{V_2} = \frac {0,450\mathrm{mol/dm^3} \cdot 0,130\mathrm{dm^3}}{0,500\mathrm{dm^3}} = 1,17\mathrm{mol/dm^3}\end{aligned}\]

c.

\[\begin{aligned} c_1V_1 &= c_2V_2 \\ V_1 &= \frac {c_2V_2}{c_1} = \frac {0,100\mathrm{mol/dm^3} \cdot 0,100\mathrm{dm^3}}{0,150\mathrm{mol/dm^3}} = 0,06666667\mathrm{dm^3} ≈ 0,0667\mathrm{dm^3}\end{aligned}\]

F53

\[\text{relativt utbyte} = \frac {\text{faktiskt utbyte}}{\text{teoretiskt utbyte}} = \frac {92,1\mathrm{g}}{106\mathrm{g}} = 0,86886792 ≈ 86,9\%\]

F54

a.

\[n_\mathrm{Fe} = \frac {m_\mathrm{Fe}}{M_\mathrm{Fe}} = \frac {250\mathrm{g}}{55,85\frac {\mathrm{g}}{\mathrm{mol}}} = 4,47627574\mathrm{mol} ≈ 4,48\mathrm{mol}\]

b.

\[\begin{aligned} n_\mathrm{FeCl_3}:n_\mathrm{Fe} &= 1:1 \\ n_\mathrm{FeCl_3} &= n_\mathrm{Fe} = 4,47627574\mathrm{mol} \\ m_\mathrm{FeCl_3} &= M_\mathrm{FeCl_3} \cdot n_\mathrm{FeCl_3} = (55,85+35,45 \cdot 3)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 4,47627574\mathrm{mol} = \\ &= 726,051925\mathrm{g} ≈ 726\mathrm{g} \end{aligned}\]

c.

\[\begin{aligned} \text{relativt utbyte} &= \frac {\text{faktiskt utbyte}}{\text{teoretiskt utbyte}} = \frac {632\mathrm{g}}{726,051925\mathrm{g}} = 0,87046116 ≈ \\ &≈ 87,0\% \end{aligned}\]

F55

  Cr2O3 + 2Al → Al2O3 + 2Cr
\[m\]  ①\[2,5\mathrm{ton}\]     ④\[1 710 526,32\mathrm{g}\]
\[n\] ②\[16 447,3684\mathrm{mol}\]     ③\[32 894,7378\mathrm{mol}\]

① \(m_\mathrm{Cr_2O_3} = 2,5\mathrm{ton} = 2,5 \cdot 10^6\mathrm{g}\)

\(\begin{aligned}n_\mathrm{Cr_2O_3} = \frac {m_\mathrm{Cr_2O_3}}{M_\mathrm{Cr_2O_3}} = \frac {2,5 \cdot 10^6\mathrm{g}}{(52,00 \cdot 2 + 16,00 \cdot 3)\mathrm{g/mol}} = 16447,3684\mathrm{mol}\end{aligned}\)

\(n_\mathrm{Cr_2O_3}:n_\mathrm{Cr} = 1:2 \\ n_\mathrm{Cr} = 2n_\mathrm{Cr_2O_3} = 2 \cdot 16447,3684\mathrm{mol} = 32894,7368\mathrm{mol}\)

\(\begin{aligned}m_\mathrm{Cr} &= M_\mathrm{Cr} \cdot n_\mathrm{Cr} = 52,00\mathrm{g/mol} \cdot 32894,7368\mathrm{mol} = \\ &= 1710526,32\mathrm{g} = \text{teoretiskt utbyte}\end{aligned}\)

Vi beräknar det faktiska utbytet:

\[\begin{aligned}\text{relativt utbyte} &= \frac {\text{faktiskt utbyte}}{\text{teoretiskt utbyte}} \\ \text{faktiskt utbyte} &= \text{relativt utbyte} \cdot \text{teoretiskt utbyte} = 0,92 \cdot 1710526,32\mathrm{g} = \\ &= 1573684,21\mathrm{g} = 1,57368421 \cdot 10^6\mathrm{g} ≈ 1,6\mathrm{ton}\end{aligned}\]

F56

  CaCO3 → CaO + CO2
\[m\] ④\[178,477175\mathrm{g}\] ①\[100\mathrm{g}\]  
\[n\] ③\[1,78316690\mathrm{mol}\] ②\[1,78316690\mathrm{mol}\]  

① \(m_\mathrm{CaO} = 100\mathrm{g}\)

② \(n_\mathrm{CaO} = \frac {m_\mathrm{CaO}}{M_\mathrm{CaO}} = \frac {100\mathrm{g}}{(40,08+16,00)\frac {\mathrm{g}}{\mathrm{mol}}} = 1,78316690\mathrm{mol}\)

③ \(n_\mathrm{CaCO_3}:n_\mathrm{CaO} = 1:1 \Rightarrow n_\mathrm{CaCO_3} = n_\mathrm{CaO} = 1,78316690\mathrm{mol}\)

\(\begin{aligned} m_\mathrm{CaCO_3} &= M_\mathrm{CaCO_3} \cdot n_\mathrm{CaCO_3} = \\ &= (40,08 + 12,01 + 16,00 \cdot 3)\mathrm{g/mol} \cdot 1,78316690\mathrm{mol} = \\ &= 178,477175\mathrm{g}\end{aligned}\)

Den massa CaCO3 som faktist behövs beräknas:

\[m_\mathrm{CaCO_3,behov} = \frac {m_\mathrm{CaCO_3}}{\text{relativt utbyte}} = \frac {178,477175\mathrm{g}}{0,79} = 225,920475\mathrm{g} ≈ 0,23\mathrm{kg}\]

Svar: Det behövs 0,23kg kalciumkarbonat.

F59

3MnO2 + 4Al 2Al2O3 + 3Mn

\[n_\mathrm{MnO_2} = \frac {m_\mathrm{MnO_2}}{M_\mathrm{MnO_2}} = \frac {63,8\mathrm{g}}{(54,94+16,00 \cdot 2)\frac {\mathrm{g}}{\mathrm{mol}}} = 0,73383943\mathrm{mol}\]\[n_\mathrm{Al} = \frac {m_\mathrm{Al}}{M_\mathrm{Al}}= \frac {25,5\mathrm{g}}{26,98\frac {\mathrm{g}}{\mathrm{mol}}}= 0,94514455\mathrm{mol}\]\[\frac {n_\mathrm{Al}}{n_\mathrm{MnO_2}} = \frac {0,94514455\mathrm{mol}}{0,73383943\mathrm{mol}}= 1,28794463< \frac {4}{3}\]

\(\Rightarrow n_\mathrm{Al}\) är begränsande.

 

3MnO2

+ 4Al

2Al2O3

+ 3Mn

\[m\]   ①\[25,5\mathrm{g}\] ④\[38,9\mathrm{g}\]  
\[n\]   ②\[0,94514455\mathrm{mol}\] ③\[0,70885841\mathrm{mol}\]  

① \(m_\mathrm{Al} = 25,5\mathrm{g}\)

② \(n_\mathrm{Al}= 0,94514455\mathrm{mol}\)

③ \(n_\mathrm{Al}:n_\mathrm{Mn} = 4:3 \\ n_\mathrm{Mn} = \frac {3}{4} n_\mathrm{Al} = \frac {3}{4} \cdot 0,94514455\mathrm{mol} = 0,70885841\mathrm{mol}\)

④ \(m_\mathrm{Mn} = M_\mathrm{Mn} \cdot n_\mathrm{Mn} = 54,94\mathrm{g/mol} \cdot 0,70885841\mathrm{mol} = 38,9446812\mathrm{g} ≈ 38,9\mathrm{g}\)

Svar: Det kan bildas maximalt 38,9 g mangan.

F60

2C(s) + O2(g) 2CO(g)

\[n_\mathrm{C} = \frac {m_\mathrm{C}}{M_\mathrm{C}} = \frac {20\mathrm{g}}{12,01\frac {\mathrm{g}}{\mathrm{mol}}} = 1,66527893\mathrm{mol} \\ n_\mathrm{C}:n_\mathrm{CO} = 1:1 \\ n_\mathrm{CO} = n_\mathrm{C} = 1,66527893\mathrm{mol}\]

\[n_\mathrm{Fe_2O_3} = \frac {m_\mathrm{Fe_2O_3}}{M_\mathrm{Fe_2O_3}} = \frac {2000\mathrm{g}}{(55,85 \cdot 2 + 16,00 \cdot 3)\mathrm{g/mol}} = 12,5234815\mathrm{mol}\]

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

\[n_\mathrm{Fe_2O_3}:n_\mathrm{CO} = 1:3 \\ \frac {n_\mathrm{Fe_2O_3}}{n_\mathrm{CO}} = \frac {12,5234815\mathrm{mol}}{1,66527893\mathrm{mol}} = 7,52035066 > \frac {1}{3}\]

\(n_\mathrm{CO}\) är begränsande.

\[n_\mathrm{CO}:n_\mathrm{Fe} = 3:2 \\ n_\mathrm{Fe} = \frac {2}{3}n_\mathrm{CO} = \frac {2}{3} \cdot 1,66527893 = 1,11018596\mathrm{mol} = 62,0038857\mathrm{g} ≈ 62\mathrm{g}\]