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, F30, F31, F32, F33, F34, F35, F36, F37, F40, F42, F43, F44, F45, F46, F47, F48, F49, F50, F51, F52, F53, F54, F55, F56, F59, F60
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
\[m = \rho \cdot V =11,3 \frac {\mathrm{g}}{\mathrm{cm^3}} \cdot 15,0\mathrm{cm^3} = 169,5\mathrm{g} ≈ 170\mathrm{g}\]
Svar: \(m = 170\mathrm{g}\).
Formeln för kaliumjodid är KI.
\[\text{procenthalt} = \frac {m_\mathrm{1K-atom}}{m_\mathrm{1KI-enhet}} = \frac {39,10\mathrm{u}}{(39,10+126,9)\mathrm{u}} = 0,23554217 ≈ 23,55\%\]
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:
\[2,00\mathrm{g} \cdot 0,25533032 = 0,51066064\mathrm{g} ≈ 0,511\mathrm{g}\]
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.
\[N_\mathrm{H} = 2N_\mathrm{H_2O} = 2 \cdot 3,37232 \cdot 10^{22} = 6,74464 \cdot 10^{22} ≈ 6,7 \cdot 10^{22}\]
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: \(m_\mathrm{CuSO_4} = 0,83\mathrm{kg}\)
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: \(n_\mathrm{Fe_3O_4} = 8,6 \cdot 10^3\mathrm{mol}\)
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 \(n_\mathrm{O_2} = 3n_\mathrm{C_2H_5OH} = 3 \cdot 1,25\mathrm{mol} = 3,75\mathrm{mol}\).
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 \[\begin{aligned}m_\mathrm{O_2} &= M_\mathrm{O_2} \cdot n_\mathrm{O_2} = (16,00 \cdot 2)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 0,04970939\mathrm{mol} = \\ &= 1,59070052\mathrm{g} ≈ 1,6\mathrm{g}\end{aligned}\]
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 \[\begin{aligned}m_\mathrm{CO_2} &= M_\mathrm{CO_2} \cdot n_\mathrm{CO_2} = (12,01 + 16,00 \cdot 2)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 1,8125\mathrm{mol} \\ &= 79,768125\mathrm{g} ≈ 80\mathrm{g}\end{aligned}\]
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
\(n_\mathrm{Fe_2O_3}:n_\mathrm{Fe} = 1:2 \\ n_\mathrm{Fe} = 2n_\mathrm{Fe_2O_3} = 2 \cdot 3130,87038\mathrm{mol} = 6261,74076\mathrm{mol}\)
④
\[\begin{aligned} m_\mathrm{Fe} &= M_\mathrm{Fe} \cdot n_\mathrm{Fe} = 55,85\frac {\mathrm{g}}{\mathrm{mol}} \cdot 6261,74076\mathrm{mol} = 349718,222\mathrm{g} = \\ &= 349,718222 \cdot 10^3\mathrm{g} ≈ 350\mathrm{kg}\end{aligned}\]
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
\(n_\mathrm{Fe}:n_\mathrm{CO_2} = 4:3 \\ n_\mathrm{CO_2} = \frac {3}{4}n_\mathrm{Fe} = \frac {3}{4} \cdot 1,02417189 \cdot 10^7\mathrm{mol} = 7,68128917 \cdot 10^6\mathrm{mol}\)
④
\(\begin{aligned}m_\mathrm{CO_2} &= M_\mathrm{CO_2} \cdot n_\mathrm{CO_2} = (12,01 + 16,00 \cdot 2)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 7,68128917 \cdot 10^6\mathrm{mol} = \\ &= 338,053536 \cdot 10^6\mathrm{g} ≈ 338\mathrm{ton}\end{aligned}\)
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 \(m_\mathrm{CoCl_2 \cdot 6H_2O} = 1,0\mathrm{g}\)
②
\(\begin{aligned}n_\mathrm{CoCl_2 \cdot 6H_2O} &= \frac {m_\mathrm{CoCl_2 \cdot 6H_2O}}{M_\mathrm{CoCl_2 \cdot 6H_2O}} = \frac {1,0\mathrm{g}}{(58,93+25,45 \cdot 2 + 6 \cdot (1,008 \cdot 2 + 16,00))\mathrm{g/mol}} = \\ &= 0,00420299\mathrm{mol}\end{aligned}\)
③
\(n_\mathrm{CoCl_2 \cdot 6H_2O}:n_\mathrm{H_2O} = 1:6 \\ n_\mathrm{H_2O} = 6n_\mathrm{CoCl_2 \cdot 6H_2O} = 6 \cdot 0,00420299\mathrm{mol} = 0,02521792\mathrm{mol}\)
④
\(\begin{aligned}m_\mathrm{H_2O} &= M_\mathrm{H_2O} \cdot n_\mathrm{H_2O} = (1,008 \cdot 2 + 16,00)\mathrm{g/mol} \cdot 0,02521792\mathrm{mol} = \\ &= 0,45432614\mathrm{g} ≈ 0,45\mathrm{g}\end{aligned}\)
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)
FeSO4 · 7H2O(s) \(\overset{\Delta}{\rightarrow}\) FeSO4(s) + 7H2O(g)
\[\begin{aligned} n_\mathrm{FeSO_4 \cdot 7H_2O} &= \frac {m_\mathrm{FeSO_4 \cdot 7H_2O}}{M_\mathrm{FeSO_4 \cdot 7H_2O}} = \\ &= \frac {1,26\mathrm{g}}{(55,85+32,07+16,00\cdot 4+7\cdot (1,008\cdot 2+16,00))\mathrm{g/mol}} = \\ &= 0,00453185\mathrm{mol}\end{aligned}\]
\(n_\mathrm{FeSO_4 \cdot 7H_2O}:n_\mathrm{FeSO_4} = 1:1\)
\(n_\mathrm{FeSO_4} = n_\mathrm{FeSO_4 \cdot 7H_2O} = 0,00453185\mathrm{mol}\)
\[\begin{aligned} m_\mathrm{FeSO_4} &= M_\mathrm{FeSO_4} \cdot n_\mathrm{FeSO_4} = \\ &= (55,85+32,07+16,00\cdot 4)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 0,00453185\mathrm{mol} = \\ &= 0,68847902\mathrm{g} ≈ 0,688\mathrm{g}\end{aligned}\]
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.
\(n_\mathrm{Cu} = \frac {m_\mathrm{Cu}}{M_\mathrm{Cu}} = \frac {0,252\mathrm{g}}{63,55\mathrm{g/mol}} = 0,00396538\mathrm{mol}\)
\(n_\mathrm{S} = \frac {m_\mathrm{S}}{M_\mathrm{S}}\)
\(m_\mathrm{S} = m_\mathrm{kopparsulfid} - m_\mathrm{Cu} = = 0,316\mathrm{g} - 0,252\mathrm{g} = 0,064\mathrm{g}\)
\(n_\mathrm{S} = \frac {m_\mathrm{S}}{M_\mathrm{S}} = \frac{0,064\mathrm{g}}{32,07\mathrm{g/mol}} = 0,00199563\mathrm{mol}\)
\(\frac{n_\mathrm{Cu}}{n_\mathrm{S}} = \frac{0,00396538\mathrm{mol}}{0,00199563\mathrm{mol}} = 1,98702793 ≈ \frac{2}{1}\)
\(n_\mathrm{Cu}:n_\mathrm{S} = 2:1\)
Svar: Det är dubbelt så många Cu som S i en kopparsulfidenhet. Därför blir kopparsulfidens formel Cu2S.
- \(m_\mathrm{Pb} = 86,6\% \cdot 100\mathrm{g}\)
\(m_\mathrm{O} = 13,4\% \cdot 100\mathrm{g}\) - \(n_\mathrm{Pb} = \frac {m_\mathrm{Pb}}{M_\mathrm{Pb}} = \frac {86,6\mathrm{g}}{207,2\mathrm{g/mol}} = 0,41795367\mathrm{mol} ≈ 0,418\mathrm{mol}\)
\(n_\mathrm{O} = \frac {m_\mathrm{O}}{M_\mathrm{O}} = \frac {13,4\mathrm{g}}{16,00\mathrm{g/mol}} = 0,8375\mathrm{mol} ≈ 0,838\mathrm{mol}\) - \(\frac {n_\mathrm{Pb}}{n_\mathrm{O}} = \frac {0,41795367\mathrm{mol}}{0,8375\mathrm{mol}} ≈ \frac {0,4\mathrm{mol}}{0,8\mathrm{mol}} = \frac {1}{2}\)
\(n_\mathrm{Pb}:n_\mathrm{O} = 1:2\)
Svar: Blyoxidens formel är PbO2.
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).
\[M_\mathrm{CH} = (12,01 + 1,008)\mathrm{g/mol} = 13,018\mathrm{g/mol} \\ \frac {M_\mathrm{faktisk}}{M_\mathrm{CH}} = \frac {78,1\mathrm{g/mol}}{13,018\mathrm{g/mol}} = \frac {M_\mathrm{faktisk}}{M_\mathrm{CH}} = \frac {78,1\mathrm{g/mol}}{13,018\mathrm{g/mol}} = 5,99938547 ≈ 6\]
Ämnets molekylformel: (CH)6 ⇔ C6H6
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.
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.
\(c_\mathrm{NaCl} = \frac {n_\mathrm{NaCl}}{V}\)
\[n_\mathrm{NaCl} = \frac {m_\mathrm{NaCl}}{M_\mathrm{NaCl}} = \frac {22\mathrm{g}}{(22,99+35,45)\frac {\mathrm{g}}{\mathrm{mol}}} = 0,37645448\mathrm{mol}\]
\(c_\mathrm{NaCl} = \frac {0,37645448\mathrm{mol}}{3,0\mathrm{dm^3}} = 0,12548483\mathrm{mol/dm^3} ≈ 0,13\mathrm{M}\)
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.
\[\begin{aligned} n_\mathrm{C_6H_{12}O_6} &= \frac {m_\mathrm{C_6H_{12}O_6}}{M_\mathrm{C_6H_{12}O_6}} = \frac {0,045\mathrm{g}}{(12,01 \cdot 6 + 1,008 \cdot 12 + 16,00 \cdot 6)\mathrm{g/mol}} =\\ &= 0,00024978\mathrm{mol} \\ c_\mathrm{C_6H_{12}O_6} &= \frac {n_\mathrm{C_6H_{12}O_6}}{V} = \frac {0,00024978\mathrm{mol}}{0,016\mathrm{dm^3}} = 0,01561147\mathrm{mol/dm^3} ≈ 0,016\mathrm{M}\end{aligned}\]
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 \([\mathrm{CuCl_2}] = 0\mathrm{mol/dm^3}\).
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.
\[\begin{aligned} c_\mathrm{MgCl_2} &= \frac {m_\mathrm{MgCl_2}}{V} = \frac {M_\mathrm{MgCl_2} \cdot n_\mathrm{MgCl_2}}{V} = \\ &= \frac {(24,31+35,45 \cdot 2)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 0,125\mathrm{mol}}{1\mathrm{dm^3}} = 11,90125\mathrm{g/dm^3} ≈ 11,9\mathrm{g/dm^3} \end{aligned}\]
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)
\(n_\mathrm{NH_4^+}:n_\mathrm{(NH_4)_2CO_3} = 2:1\)
\[\begin{aligned}\mathrm{[NH_4^+]} &= 2c_\mathrm{(NH_4)_2CO_3} = 2 \cdot 1,56 \mathrm{mol/dm^3} = 3,12\mathrm{mol/dm^3} \\ V &= \frac {n_\mathrm{NH_4^+}}{[\mathrm{NH_4^+}]} = \frac {0,75\mathrm{mol}}{3,12\mathrm{mol/dm^3}} = 0,24038462\mathrm{dm^3} ≈ 0,24\mathrm{dm^3}\end{aligned}\]
\[c = \frac {m_\mathrm{paracetamol}}{m_\mathrm{tot}} = \frac {2 \cdot 500 \cdot 10^{-3}\mathrm{g}}{(93000 + 2 \cdot 500 \cdot 10^{-3}\mathrm{g})} = 1,075266 \cdot 10^{-5} ≈ 0,0011\%\]
\[[\mathrm{Cl^-}] = \frac {n_\mathrm{Cl^-,tot}}{V} = \frac {n_\mathrm{Cl^-,NaCl} + n_\mathrm{Cl^-,MgCl_2}}{V}\]
\[\begin{aligned}n_\mathrm{Cl^-,NaCl} &= n_\mathrm{NaCl} = \frac {m_\mathrm{NaCl}}{M_\mathrm{NaCl}} = \frac {1,00\mathrm{g}}{(22,99+35,45)\mathrm{g/mol}} = 0,01711157\mathrm{mol} \\ n_\mathrm{Cl^-,MgCl_2} &= 2n_\mathrm{MgCl_2} = 2\frac {m_\mathrm{MgCl_2}}{M_\mathrm{MgCl_2}} = 2 \cdot \frac {1,00\mathrm{g}}{(24,31+35,45 \cdot 2)\mathrm{g/mol}} = 0,01329610\mathrm{mol}\end{aligned}\]
\[[\mathrm{Cl^-}] = \frac {(0,01711157 + 0,01329610)\mathrm{mol}}{0,100\mathrm{dm}^3} = 0,38117764\text{mol/dm}^3 ≈ 0,381\mathrm{mol/dm^3}\]
\[\begin{aligned} c = \frac {n_\mathrm{NaOH,tot}}{V_\mathrm{tot}} &= \frac {n_\mathrm{NaOH,A} + n_\mathrm{NaOH,B}}{V_\mathrm{A} + V_\mathrm{B}} = \\ &= \frac {c_\mathrm{NaOH,A} \cdot V_\mathrm{A} + c_\mathrm{NaOH,B} \cdot V_\mathrm{B}}{V_\mathrm{A} + V_\mathrm{B}} = \\ &= \frac {2,5\frac {\mathrm{mol}}{\mathrm{dm}^3} \cdot 0,030\mathrm{dm}^3 + 0,50\frac {\mathrm{mol}}{\mathrm{dm}^3} \cdot 0,075\mathrm{dm}^3}{(0,030 + 0,075)\mathrm{dm^3}} = \\ & = 1,07142857\mathrm{mol/dm^3} ≈ 1,1\mathrm{mol/dm^3} \end{aligned}\]
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:
\[\begin{aligned} \mathrm{[Cl^-]} &= \frac {0,00176491\mathrm{mol}}{0,250\mathrm{dm^3}} = 0,00705964\mathrm{mol/dm^3} ≈ \\ &≈ 0,00706\mathrm{mol/dm^3} = 7,06 \cdot 10^{-3}\mathrm{mol/dm^3}\end{aligned}\]
\[\begin{aligned} c_1V_1 &= c_2V_2 \\ V_1 &= \frac {c_2V_2}{c_1} = \frac {2,5\mathrm{mol/dm^3} \cdot 5,0\mathrm{dm^3}}{11,8\mathrm{mol/dm^3}} = 1,05932203\mathrm{dm^3} ≈ 1,1\mathrm{dm^3}\end{aligned}\]
\[\begin{aligned} c_1V_1 &= c_2V_2 \\ V_2 &= \frac {c_1V_1}{c_2} = \frac {10\% \cdot 15\mathrm{ml}}{3,0\%} = 50\mathrm{ml}\end{aligned}\]
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.
\[\begin{aligned} c_1V_1 &= c_2V_2 \\ V_1 &= \frac {c_2V_2}{c_1} = \frac {0,100\mathrm{mol/dm^3} \cdot 0,100\mathrm{dm^3}}{0,150\mathrm{mol/dm^3}} = 0,06666667\mathrm{dm^3} ≈ 0,0667\mathrm{dm^3}\end{aligned}\]
\[\text{relativt utbyte} = \frac {\text{faktiskt utbyte}}{\text{teoretiskt utbyte}} = \frac {92,1\mathrm{g}}{106\mathrm{g}} = 0,86886792 ≈ 86,9\%\]
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.
\[\begin{aligned} \text{relativt utbyte} &= \frac {\text{faktiskt utbyte}}{\text{teoretiskt utbyte}} = \frac {632\mathrm{g}}{726,051925\mathrm{g}} = 0,87046116 ≈ \\ &≈ 87,0\% \end{aligned}\]
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 \(m_\mathrm{Cr_2O_3} = 2,5\mathrm{ton} = 2,5 \cdot 10^6\mathrm{g}\)
②
\(\begin{aligned}n_\mathrm{Cr_2O_3} = \frac {m_\mathrm{Cr_2O_3}}{M_\mathrm{Cr_2O_3}} = \frac {2,5 \cdot 10^6\mathrm{g}}{(52,00 \cdot 2 + 16,00 \cdot 3)\mathrm{g/mol}} = 16447,3684\mathrm{mol}\end{aligned}\)
③
\(n_\mathrm{Cr_2O_3}:n_\mathrm{Cr} = 1:2 \\ n_\mathrm{Cr} = 2n_\mathrm{Cr_2O_3} = 2 \cdot 16447,3684\mathrm{mol} = 32894,7368\mathrm{mol}\)
④
\(\begin{aligned}m_\mathrm{Cr} &= M_\mathrm{Cr} \cdot n_\mathrm{Cr} = 52,00\mathrm{g/mol} \cdot 32894,7368\mathrm{mol} = \\ &= 1710526,32\mathrm{g} = \text{teoretiskt utbyte}\end{aligned}\)
Vi beräknar det faktiska utbytet:
\[\begin{aligned}\text{relativt utbyte} &= \frac {\text{faktiskt utbyte}}{\text{teoretiskt utbyte}} \\ \text{faktiskt utbyte} &= \text{relativt utbyte} \cdot \text{teoretiskt utbyte} = 0,92 \cdot 1710526,32\mathrm{g} = \\ &= 1573684,21\mathrm{g} = 1,57368421 \cdot 10^6\mathrm{g} ≈ 1,6\mathrm{ton}\end{aligned}\]
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 \(n_\mathrm{CaCO_3}:n_\mathrm{CaO} = 1:1 \Rightarrow n_\mathrm{CaCO_3} = n_\mathrm{CaO} = 1,78316690\mathrm{mol}\)
④
\(\begin{aligned} m_\mathrm{CaCO_3} &= M_\mathrm{CaCO_3} \cdot n_\mathrm{CaCO_3} = \\ &= (40,08 + 12,01 + 16,00 \cdot 3)\mathrm{g/mol} \cdot 1,78316690\mathrm{mol} = \\ &= 178,477175\mathrm{g}\end{aligned}\)
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.
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.
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.
\[n_\mathrm{CO}:n_\mathrm{Fe} = 3:2 \\ n_\mathrm{Fe} = \frac {2}{3}n_\mathrm{CO} = \frac {2}{3} \cdot 1,66527893 = 1,11018596\mathrm{mol} = 62,0038857\mathrm{g} ≈ 62\mathrm{g}\]