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, 5.31, 5.32, 5.35, 5.37, 5.39, 5.43

5.3.

[Video kommer i augusti–september 2021.]

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

5.31

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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