Magnus Ehingers undervisning

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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 \[\begin{aligned}m_\mathrm{CO_2} &= M_\mathrm{CO_2} \cdot n_\mathrm{CO_2} = (12,01 + 16,00 \cdot 2)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 1,8125\mathrm{mol} \\ &= 79,768125\mathrm{g} ≈ 80\mathrm{g}\end{aligned}\]