Magnus Ehingers undervisning

Allt du behöver för A i Biologi, Kemi, Bioteknik, Gymnasiearbete m.m.

Kemi 1

Administration

Prov 2011-09-21 i Kemiska beräkningar

Artikelindex

Facit

Betygsgränser

Max: 15
G: 5
VG: 9

Provets omfattning är för liten för att man ska kunna sätta några högre betyg.

Del I. Fullständiga lösningar ska redovisas.

Ingen/felaktig enhet i svaret … -1p

Räknat med avrundade siffror … -0,5p

  1. \(m = Mn = (23,0+35,5)\text{g/mol} \cdot 1,5\text{mol} = 87,75\text{g} \approx 88\text{g}\)
    Ej räknat med mtot utan bara med mvatten … -0,5p

    1. \(c = \frac {m_{\text{socker}}}{m_{\text{tot}}} = \frac {10{\text{g}}}{(10+100){\text{g}}} = 0,09090909 \approx 9,1\%\)
    2. \(c = \frac {n}{V}\)
      \(n = \frac {m}{M} = \frac {10\text{g}}{(12,0·12+1,008·22+16,0·11)\text{g/mol}} = 0,02922473\text{mol}\)
      \(V = 100\text{cm}^3 = 0,100\text{dm}^3\)

      \(c = \frac {0,02922473\text{mol}}{0,100\text{dm}^3} = 0,29224726\text{mol/dm}^3 \approx 0,29\text{M}\)

  2. \(pV = nRT \Leftrightarrow n = \frac {pV}{RT}\) och \(m=Mn \Leftrightarrow n= \frac{m}{M}\)
    Vi får då att \(\frac {m}{M} = \frac {pV}{RT} \Leftrightarrow m = \frac {MpV}{RT}\)

    \(\\ p = 102,4k\text{Pa} = 102400 \text{N/m}^2 \\ V = 9,5\text{L} = 0,0095\text{m}^3 \\ M = 4,0 \text{g/mol} \\ R = 8,314 \frac {\text{Nm}}{\text{molK}} \\ T = 13,5^{\circ}\text{C} = (273,15 + 13,5)\text{K} = 286,65\text{K}\)

    \(m = \frac {102400\frac{\text{N}}{\text{m}^2} \cdot 0,0095\text{m}^3 \cdot 4,0\frac{\text{g}}{\text{mol}}}{8,314\frac{\text{Nm}}{\text{molK}} \cdot 286,65\text{K}} = 1,6327571\text{g} \approx 1,6\text{g}\)

  3. \(\\m_{\text{Na}_2\text{CO}_3} = M_{\text{Na}_2\text{CO}_3} \cdot n_{\text{Na}_2\text{CO}_3} = M_{\text{Na}_2\text{CO}_3} \cdot \frac {1}{2}n_{\text{NaHCO}_3} = M_{\text{Na}_2\text{CO}_3} \cdot \frac {1}{2} \cdot \frac {m_{\text{NaHCO}_3}}{M_{\text{NaHCO}_3}} =\)
    \(= (23,0 \cdot 2 + 12,0 + 16,0 \cdot 3)\text{g/mol} \cdot \frac {1}{2} \cdot \frac {500\text{g}}{(23,0 + 1,008 + 12,0 + 16,0 \cdot 3)\text{g/mol}} =\)
    \(= 315,446148\text{g} \approx 0,315k\text{g}\)
  4. \(n_{\text{Fe}} = \frac {m_{\text{Fe}}}{M_{\text{Fe}}} = \frac {5,0\text{g}}{55,8\text{g/mol}} = 0,08960573\text{mol}\)

    \(n_{\text{S}} = \frac {m_{\text{S}}}{M_{\text{S}}} = \frac {2,5\text{g}}{32,0\text{g/mol}} = 0,078125\text{mol}\)


    Eftersom \(n_{\text{S}} < n_{\text{Fe}}\) kommer \(n_{\text{S}}\) att vara begränsande. Alltså kan det bildas maximalt 0,0078125 mol FeS.


    \(m_{\text{FeS}} = n_{\text{FeS}} \cdot M_{\text{FeS}} =\)
    \(= 0,078125mol \cdot (55,8 + 32,0)\text{g/mol}= 6,859375\text{g} \approx 6,9\text{g}\)


  5. 2C8H18 + 25O2 → 18H2O + 16CO2
    \(m_{\text{CO}_2} = 12M_{\text{CO}_2} \cdot n_{\text{CO}_2} =12M_{\text{CO}_2} \cdot 8n_{\text{C}_8\text{H}_{18}}\)
    \(n_{\text{C}_8\text{H}_{18}} = \frac {m_{\text{C}_8\text{H}_{18}}}{M_{\text{C}_8\text{H}_{18}}} = \frac {\rho_{\text{C}_8\text{H}_{18}} \cdot V_{\text{C}_8\text{H}_{18}}}{M_{\text{C}_8\text{H}_{18}}}\)


    \(m_{\text{CO}_2} = 12M_{\text{CO}_2} \cdot 8 \frac {\rho_{\text{C}_8\text{H}_{18}} \cdot V_{\text{C}_8\text{H}_{18}}}{M_{\text{C}_8\text{H}_{18}}} =\)
    \(= 12 \cdot (12,0 + 16,0 \cdot 2)\text{g/mol} \cdot 8 \cdot \frac {0,703\text{g/L} \cdot 80\text{L}}{(12,0 \cdot 8 + 1,008 \cdot 18)\text{g/mol}} =\)
    \(= 2081211,1\text{g} \approx 2,1\text{ton}\)

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