Magnus Ehingers undervisning

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Kemi 1

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Kapitel 5

kapitel 5Kompletta lösningar (med videor) till utvalda övningsuppgifter i kapitel 5 i Ehinger: Kemi 1 (NA förlag).

Hoppa direkt till uppgift: 5.3, 5.5, 5.8, 5.10, 5.12–13, 5.14, 5.21, 5.22, 5.31, 5.32, 5.35, 5.37, 5.39, 5.43

5.3.

\[\rho = \frac {m}{V} = \frac {154\mathrm{g}}{8\mathrm{cm^3}} = 19,25\mathrm{g/cm^3} \approx 19\mathrm{g/cm^3} \hspace{100cm}\]

5.5

\[\begin{aligned} m_\mathrm{Li} &= 7,42\% \cdot m_{^6\mathrm{Li}} + 92,58\% \cdot m_{^7\mathrm{Li}} = \hspace{100cm} \\ &= 0,0742 \cdot 6,0151\mathrm{u} + 0,9258 \cdot 7,0160\mathrm{u} = \\ &= 6,94173322\mathrm{u} \approx 6,94\mathrm{u}\end{aligned}\]

5.8

\[\mathrm{mass\%} = \frac {m_\mathrm{Cl-atom}}{m_\mathrm{NaCl-enhet}} = \frac {35,5\mathrm{u}}{(23,0 + 35,5)\mathrm{u}} = 0,60683761 \approx 60,7\% \hspace{100cm}\]

5.10

\[\begin{aligned} N_\mathrm{NH_3} &= n_\mathrm{NH_3} \cdot N_\mathrm{A} = 0,050\mathrm{mol} \cdot 6,022 \cdot 10^{23}/\mathrm{mol} = \hspace{100cm} \\ &= 3,011 \cdot 10^{22} \approx 3,0 \cdot 10^{22}\end{aligned}\]

\[N_\mathrm{N} = N_\mathrm{NH_3} = 3,0 \cdot 10^{22} \hspace{100cm}\]

\[N_\mathrm{H} = 3N_\mathrm{NH_3} = 3 \cdot 3,0 \cdot 10^{22} = 9,0 \cdot 10^{22} \hspace{100cm}\]

5.12–13

a.

\[m_\mathrm{Cu} = 10\mathrm{g} \hspace{100cm}\]

\[M_\mathrm{Cu} = 63,5\mathrm{g/mol} \hspace{100cm}\]

\[n_\mathrm{Cu} = \frac {m_\mathrm{Cu}}{M_\mathrm{Cu}} = \frac {10\mathrm{g}}{63,5\mathrm{g/mol}} = 0,15748031\mathrm{mol} \approx 0,16\mathrm{mol}\hspace{100cm}\]

b.

\[m_\mathrm{CuSO_4} = 10\mathrm{g} \hspace{100cm}\]

\[\begin{aligned} M_\mathrm{CuSO_4} &= M_\mathrm{Cu} + M_\mathrm{S} + 4M_\mathrm{O} = \hspace{100cm} \\ &= (63,5+32,1+4 \cdot 16,0)\mathrm{g/mol} = 159,6\mathrm{g/mol}\end{aligned}\]

\[n_\mathrm{CuSO_4} = \frac {m_\mathrm{CuSO_4}}{M_\mathrm{CuSO_4}} = \frac {10\mathrm{g}}{159,6\mathrm{g/mol}} = 0,06265664\mathrm{mol} \approx 0,063\mathrm{mol} \hspace{100cm}\]

c.

\[m_\mathrm{CO_2} = 10\mathrm{g} \hspace{100cm}\]

\[M_\mathrm{CO_2} = M_\mathrm{C} + 2M_\mathrm{O} = (12,0 + 2 \cdot 16,0)\mathrm{g/mol} = 44,0\mathrm{g/mol} \hspace{100cm}\]

\[n_\mathrm{CO_2} = \frac {m_\mathrm{CO_2}}{M_\mathrm{CO_2}} = \frac {10\mathrm{g}}{44,0\mathrm{g/mol}} = 0,22727273\mathrm{mol} \approx 0,23\mathrm{mol} \hspace{100cm}\]

d.

\[m_\mathrm{HCl} = 10\mathrm{g} \hspace{100cm}\]

\[M_\mathrm{HCl} = M_\mathrm{H} + M_\mathrm{Cl} = (1,01 + 35,5)\mathrm{g/mol} = 36,51\mathrm{g/mol} \hspace{100cm}\]

\[n_\mathrm{HCl} = \frac {m_\mathrm{HCl}}{M_\mathrm{HCl}} = \frac {10\mathrm{g}}{36,51\mathrm{g/mol}} = 0,27389756\mathrm{mol} \approx 0,27\mathrm{mol} \hspace{100cm}\]

e.

\[m_\mathrm{CH_3OH} = 10\mathrm{g} \hspace{100cm}\]

\[M_\mathrm{CH_3OH} = M_\mathrm{C} + 4M_\mathrm{H} + \mathrm{O} = (12,0 + 4 \cdot 1,01 + 16,0)\mathrm{g/mol} = 32,04\mathrm{g/mol} \hspace{100cm}\]

\[n_\mathrm{CH_3OH} = \frac {m_\mathrm{CH_3OH}}{M_\mathrm{CH_3OH}} = \frac {10\mathrm{g}}{32,04\mathrm{g/mol}} = 0,31210986\mathrm{mol} \approx 0,31\mathrm{mol} \hspace{100cm}\]

5.14.

Al2(SO4)3 · xH2O(s) \({\sf \overset{\Delta}{\rightarrow}}\) Al2(SO4)3(s) + xH2O(g)

\[m_{\mathrm{Al_2(SO_4)_3 \cdot}x\mathrm{H_2O}} = 3,85\mathrm{g} \hspace{100cm}\]

\[m_\mathrm{Al_2(SO_4)_3} = 1,98\mathrm{g} \hspace{100cm}\]

\[m_\mathrm{H_2O} = m_{\mathrm{Al_2(SO_4)_3 \cdot}x\mathrm{H_2O}} - m_\mathrm{Al_2(SO_4)_3} = (3,85-1,98)\mathrm{g} = 1,87\mathrm{g} \hspace{100cm}\]

\[\begin{aligned} n_\mathrm{Al_2(SO_4)_3} &= \frac {m_\mathrm{Al_2(SO_4)_3}}{M_\mathrm{Al_2(SO_4)_3}} = \hspace{100cm} \\ &= \frac {1,98\mathrm{g}}{(26,98 \cdot 2 + (32,07 + 16,00 \cdot 4) \cdot 3)\mathrm{g/mol}} = 0,00578660\mathrm{mol} \end{aligned}\]

\[n_\mathrm{H_2O} = \frac {m_\mathrm{H_2O}}{M_\mathrm{H_2O}} = \frac {1,87\mathrm{g}}{(1,008 \cdot 2 + 16,00)\mathrm{g/mol}} = 0,10379663\mathrm{mol} \hspace{100cm}\]

\[x = \frac {n_\mathrm{H_2O}}{n_\mathrm{Al_2(SO_4)_3}} = \frac {0,10379663\mathrm{mol}}{0,00578660\mathrm{mol}} = 17,9374207 \approx 18 \hspace{100cm}\]

Svar: Formeln för kristalliserat aluminiumsulfat är Al2(SO4)3 · 18H2O.

5.21.

  Sn  + O2 → SnO2
\(m\) ① \[10,0 \text{g}\]   ④ \[12,7\text{g}\]
\(n\) ② \[0,08403361\text{mol}\]   ③ \[0,08403361\text{mol}\]

 

② \[n_\mathrm{Sn} = \frac {m_\mathrm{Sn}}{M_\mathrm{Sn}} = \frac {10,0\mathrm{g}}{119\mathrm{g/mol}} = 0,08403361\mathrm{mol} \hspace{100cm}\]

③ \[n_\mathrm{Sn}:n_\mathrm{SnO_2} = 1:1 \hspace{100cm}\]

\[n_\mathrm{SnO_2} = n_\mathrm{Sn} = 0,08403361\mathrm{mol} \hspace{100cm}\]

④ \[\begin{aligned} m_\mathrm{SnO_2} &= M_\mathrm{SnO_2} \cdot n_\mathrm{SnO_2} = \hspace{100cm} \\ &= (119 + 16,0 \cdot 2)\mathrm{g/mol} \cdot 0,08403361\mathrm{mol} = \\ &= 12,6890756\mathrm{g} \approx 12,7\mathrm{g} \end{aligned}\]

Svar: \(m_\mathrm{SnO_2} = 12,7\mathrm{g}\)

5.22.

  4CH3NO2
+ 5O2 → 4CO2 + 6H2O + 4NO
\(m\) ① \[200 \text{g}\]   ④ \[131\text{g}\]
\(n\) ② \[3,27707685\text{mol}\]   ③ \[4,09634605\text{mol}\]

 

② \[\begin{aligned} n_\mathrm{CH_3NO_2} &= \frac {m_\mathrm{CH_3NO_2}}{M_\mathrm{CH_3NO_2}} = \hspace{100cm} \\ &= \frac {200\mathrm{g}}{(12,0+1,01 \cdot 3 + 14,0 + 16,0 \cdot 2)\mathrm{g/mol}} = \\ &= 3,27707685\mathrm{mol} \end{aligned}\]

③ \[n_\mathrm{CH_3NO_2} : n_\mathrm{O_2} = 4:5 \hspace{100cm}\]

\[n_\mathrm{O_2} = \frac {5}{4} \cdot n_\mathrm{CH_3NO_2} = \frac {5}{4} \cdot 3,27707685\mathrm{mol}= 4,09634605\mathrm{mol}  \hspace{100cm}\]

④ \[\begin{aligned} m_\mathrm{O_2} &= M_\mathrm{O_2} \cdot n_\mathrm{O_2} = \hspace{100cm} \\ &= (16,0 \cdot 2)\mathrm{g/mol} \cdot 4,09634605\mathrm{mol} = \\ &= 131,083074\mathrm{g} \approx 12,7\mathrm{g} \approx 131\mathrm{g} \end{aligned}\]

Svar: \(m_\mathrm{O_2} = 131\mathrm{g}\)

5.31

a.

\[c = \frac {m}{V} = \frac {15\text{g}}{0,500\text{dm}^3} = 30\text{g/dm}^3 \hspace{100cm}\]

b.

\[c = \frac {m_{\text{MgCl}_2}}{m_\text{tot}} = \frac {15\text{g}}{(500+15)\text{g}} \approx 2,9\% \hspace{100cm}\]

c.

\[c = \frac {m}{V} = \frac {15}{500} = 0,030 = 3,0\% \hspace{100cm}\]

d.

\[\begin{aligned} c_{\text{MgCl}_2} &= \frac {n_{\text{MgCl}_2}}{V} = \frac {\frac {m_{\text{MgCl}_2}}{M_{\text{MgCl}_2}}}{V} = \frac {\frac {15\text{g}}{(24,3+35,3\cdot 2)\text{g/mol}}}{0,500} = \hspace{100cm} \\ &= 0,31469538 \text{mol/dm}^3 \approx 0,31\text{mol/dm}^3 \end{aligned}\]

e.

När MgCl2(s) löses upp sker följande:

MgCl2(s) → Mg2+(aq) + 2Cl(aq)

Vi ser att \(n_{\text{MgCl}_2}:n_{\text{Mg}^{2+}} = 1:1\). Därför blir \(n_\text{Mg} = n_{\text{MgCl}_2}\) och vi får att

\[[\text{Mg}^{2+}] = c_{\text{MgCl}_2} = 0,31\text{mol/dm}^3 \hspace{100cm}\]

f.

När MgCl2(s) löses upp sker följande:

MgCl2(s) → Mg2+(aq) + 2Cl(aq)

Vi ser att \(n_{\text{MgCl}_2}:n_{\text{Cl}^-} = 1:2\). Därför blir \(n_{\text{Cl}^-} = 2n_{\text{MgCl}_2}\) och vi får att

\[\begin{aligned} \mathrm{[Cl^-]} &= 2c_{\text{MgCl}_2} = 2 \cdot 0,31469538 \text{mol/dm}^3 = 0,629590766\text{mol/dm}^3 = \hspace{100cm} \\ &\approx 0,63\text{mol/dm} \end{aligned}\]

g.

När MgCl2(s) löses upp sker följande:

MgCl2(s) → Mg2+(aq) + 2Cl(aq)

Eftersom allt MgCl2(s) omvandlas till Mg2+(aq) + 2Cl(aq) finns det inga MgCl2-partiklar kvar. Därför blir [MgCl2] = 0 mol/dm3.

5.32

c.

\[m_{\text{C}_6\text{H}_{12}\text{O}_6} = M_{\text{C}_6\text{H}_{12}\text{O}_6} \cdot n_{\text{C}_6\text{H}_{12}\text{O}_6} = 180,2\text{g/mol} \cdot 0,10\text{mol} = 18,02\text{g} \hspace{100cm}\]

\[c = \frac {m_{\text{C}_6\text{H}_{12}\text{O}_6}}{m_{\text{tot}}} = \frac {18,02\text{g}}{(200 + 18,02)\text{g}} = 0,08265 \approx 8,3\% \hspace{100cm}\]

d.

\[m_{\text{C}_6\text{H}_{12}\text{O}_6} = M_{\text{C}_6\text{H}_{12}\text{O}_6} \cdot n_{\text{C}_6\text{H}_{12}\text{O}_6} = 180,2\text{g/mol} \cdot 0,10\text{mol} = 18,02\text{g} \hspace{100cm}\]

\[c = \frac {m}{V} = \frac {18,02}{200} = 0,09009 \approx 9,0\% \hspace{100cm}\]

5.35

\[c_1V_1 = c_2V_2 \hspace{100cm}\]

\[V_2 = \frac {c_1V_1}{c_2} = \frac {20\% \cdot 10\text{cm}^3}{5\%} = 40\text{cm}^3 \hspace{100cm}\]

5.37

  Cr2O3 + 2Al → Al2O3 + 2Cr
\(m\) ① \[500 \text{kg}\]   ④ \[342105,2632\text{g}\]
\(n\) ② \[3289,473684\text{mol}\]   ③ \[6578,947368\text{mol}\]

 

② \[n_{\text{Cr}_2\text{O}_3} = \frac {m_{\text{Cr}_2\text{O}_3}}{M_{\text{Cr}_2\text{O}_3}} = \frac {500000\text{g}}{(52,0 \cdot 2 + 16,0 \cdot 3)\text{g/mol}} = 3289,473684\text{mol} \hspace{100cm}\]

③ \[n_{\text{Cr}_2\text{O}_3}:n_\text{Cr} = 1:2 \hspace{100cm}\]

\[n_\text{Cr} = 2n_{\text{Cr}_2\text{O}_3} = 2 \cdot 3289,473684\text{mol} = 6578,947368\text{mol} \hspace{100cm}\]

④ \[m_\text{Cr} = n_\text{Cr} \cdot M_\text{Cr} = 6578,947368\text{mol} \cdot 52,0\text{g/mol} = 342105,2632\text{g} \hspace{100cm}\]

\[\begin{aligned} \text{praktiskt utbyte} &= \text{utbyte} \cdot \text{teoretiskt utbyte} = \hspace{100cm} \\ &= 95\% \cdot 342105,2632\text{g} =\\ &= 325000\text{g} = 325\text{kg} \end{aligned}\]

5.39

a.

H2 + Br2 → 2HBr

\[n_{\text{H}_2} = \frac {m_{\text{H}_2}}{M_{\text{H}_2}} = \frac {0,202\text{g}}{1,008\text{g/mol} \cdot 2} = 0,10019841\text{mol} \hspace{100cm}\]

\[n_{\text{Br}_2} = \frac {m_{\text{Br}_2}}{M_{\text{Br}_2}} = \frac {40,0\text{g}}{79,9\text{g/mol} \cdot 2} = 0,25031289\text{mol} \hspace{100cm}\]

\[n_{\text{H}_2}:n_{\text{Br}_2} = 1:1 \hspace{100cm}\]

\(n_{\text{H}_2} < n_{\text{Br}_2} \Rightarrow n_{\text{H}_2}\) är begränsande.

b.

H2 + Br2 → 2HBr

\[n_{\text{H}_2}:n_{\text{HBr}} = 1:2 \hspace{100cm}\]

\[n_{\text{HBr}} = 2n_{\text{H}_2} = 2 \cdot 0,10019841\text{mol} = 0,20039683\text{mol} \hspace{100cm}\]

\[\begin{aligned} m_{\text{HBr}} &= n_{\text{HBr}} \cdot M_{\text{HBr}} = 0,20039683\text{mol} \cdot (1,008 + 79,9)\text{g/mol} = \hspace{100cm} \\ &= 16,2137063\text{g} \approx 16,2\text{g} \end{aligned}\]

5.43.

Massan kalkopyrit i 6 miljoner ton kopparmalm:

\[m_{\mathrm{CuFeS}_2} = 6000000\text{ton} \cdot 12000\text{g/ton} = 72 \cdot 10^9\text{g} \hspace{100cm}\]

Andel koppar i kalkopyrit:

\[\frac {m_\mathrm{Cu-atom}}{m_{\mathrm{CuFeS_2-enhet}}} = \frac {63,5\text{u}}{(63,5+55,8+32,1\cdot 2)\text{u}} = 0,34604905 \hspace{100cm}\]

Massan koppar i kalkopyriten:

\[\begin{aligned} m_\mathrm{Cu} &= 0,34604905 \cdot m_{\mathrm{CuFeS}_2} = 0,34604905 \cdot 72 \cdot 10^9\text{g} = \hspace{100cm} \\ &= 24,915531335 \cdot 10^9\mathrm{g} \approx 25 \cdot 10^3\mathrm{ton}\end{aligned}\]

 

 

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