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Facit till Läxförhör A 2007-11-22 i mol och stökiometri för Nv1A

Betygsgränser

Max: 9,5
Medel: 5,34
G: 3,0
VG: 6,0
MVG: **Kan ej sättas**

Beräkna!

    1. \[n = \frac {m}{M} = \frac {0,256\text{g}}{197,0\text{g/mol}} = 0,001299492\text{mol} \approx 1,30\text{mmol} \hspace{100cm}\]

    2.  

      \[\begin{align}N &= nN_{\text{A}} = 0,001299492\text{mol} \cdot 6,022 \cdot 10^{23} = \hspace{100cm} \\ &= 7,825543 \cdot 10^{23} \approx 7,83 \cdot 10^{23}\end{align}\]

    1. C2H5OH + 3O2 → 2CO2 + 3H2O
    2. 3 mol
    3.  

      \[\begin{align}n_{\text{C}_2\text{H}_5\text{OH}} &= \frac {m_{\text{C}_2\text{H}_5\text{OH}}}{M_{\text{C}_2\text{H}_5\text{OH}}} = \hspace{100cm} \\ &= \frac {24,5\text{g}}{(12,0 \cdot 2 +1,008 \cdot 6 + 16,0)\text{g/mol}} = \frac {24,5\text{g}}{46,048\text{g/mol}} = \\ &= 0,5320535\text{mol}\end{align}\]

      \[n_{\text{H}_2\text{O}} = 3n_{\text{C}_2\text{H}_5\text{OH}} = 3 \cdot 0,5320535\text{mol} = 1,5961605\text{mol} \hspace{100cm}\]

      \[\begin{align}m_{\text{H}_2\text{O}} &= n_{\text{H}_2\text{O}} \cdot M_{\text{H}_2\text{O}} = \hspace{100cm} \\ &= 1,5961605\text{mol} \cdot (1,008 \cdot 2 + 16,0)\text{g/mol} = \\ &= 28,756428\text{g} \approx 28,8\text{g}\end{align}\]

  3. 12p+, 12n, 10e–

Facit till Läxförhör B 2007-11-22 i mol och stökiometri för Nv1A

Betygsgränser

Max: 9,5
Medel: 5,34
G: 3,0
VG: 6,0
MVG: **Kan ej sättas**

Beräkna!

    1. \[n = \frac {m}{M} = \frac {0,256\text{g}}{195,1\text{g/mol}} = 0,0013121\text{mol} \approx 1,31\text{mmol} \hspace{100cm}\]

    2.  

      \[\begin{align}N &= nN_{\text{A}} = 0,001312148\text{mol} \cdot 6,022 \cdot 10^{23} = \hspace{100cm} \\ &= 7,901752947 \cdot 10^{23} \approx 7,90 \cdot 10^{23}\end{align}\]

    1. C3H8 + 5O2 → 3CO2 + 4H2O
    2. 4 mol
    3.  

      \[\begin{align}n_{\text{C}_3\text{H}_8} &= \frac {m_{\text{C}_3\text{H}_8}}{M_{\text{C}_3\text{H}_8}} = \frac {24,5\text{g}}{(12,0 \cdot 2 +1,008 \cdot 8)\text{g/mol}} = \hspace{100cm} \\ &= 0,5560094\text{mol}\end{align}\]

      \[n_{\text{H}_2\text{O}} = 4n_{\text{C}_3\text{H}_8} = 4 \cdot 0,5560094\text{mol} = 2,2240378\text{mol} \hspace{100cm}\]

      \[\begin{align}m_{\text{H}_2\text{O}} &= n_{\text{H}_2\text{O}} \cdot M_{\text{H}_2\text{O}} = \hspace{100cm} \\ &= 2,2240378\text{mol} \cdot (1,008 \cdot 2 + 16,0)\text{g/mol} = \\ &= 40,068264\text{g} \approx 40,1\text{g}\end{align}\]

  3. 17p+, 18n, 18e–