Question

a gas mixture contains 12.1 mol of c₂h₂ and 20.6 mol of h₂.
compute the mole fraction of c₂h₂ in the mixture.
xc₂h₂=
a catalyst is then added to the mixture, and the c₂h₂ starts to react with the h₂ to give c₂h₆:
c₂h₂(g) + 2h₂(g) → c₂h₆(g)
at a certain point in the reaction, 2.70 mol of c₂h₆ is present. determine the mole fraction of c₂h₂ in the new mixture.
xc₂h₂=

Explanation:

Step1: Calculate moles of \(C_2H_2\) consumed

From the reaction \(C_2H_2(g)+2H_2(g)\to C_2H_6(g)\), the mole ratio of \(C_2H_2\) to \(C_2H_6\) is \(1:1\). So moles of \(C_2H_2\) consumed \(n_{C_2H_2,\text{consumed}}= 2.70\space mol\)

Step2: Calculate initial moles of \(C_2H_2\)

Initial moles of \(C_2H_2,n_{C_2H_2,\text{initial}} = 12.1\space mol\)

Step3: Calculate moles of \(C_2H_2\) remaining

Moles of \(C_2H_2\) remaining \(n_{C_2H_2,\text{remaining}}=n_{C_2H_2,\text{initial}}-n_{C_2H_2,\text{consumed}}=12.1 - 2.70=9.4\space mol\)

Step4: Calculate moles of \(H_2\) consumed

From the reaction, mole ratio of \(H_2\) to \(C_2H_6\) is \(2:1\). So moles of \(H_2\) consumed \(n_{H_2,\text{consumed}}=2\times2.70 = 5.4\space mol\)

Step5: Calculate moles of \(H_2\) remaining

Initial moles of \(H_2,n_{H_2,\text{initial}} = 20.6\space mol\). Moles of \(H_2\) remaining \(n_{H_2,\text{remaining}}=n_{H_2,\text{initial}}-n_{H_2,\text{consumed}}=20.6 - 5.4=15.2\space mol\)

Step6: Calculate total moles in the new mixture

Total moles \(n_{\text{total}}=n_{C_2H_2,\text{remaining}}+n_{H_2,\text{remaining}}+n_{C_2H_6}\)
\(n_{\text{total}}=9.4 + 15.2+2.70=27.3\space mol\)

Step7: Calculate mole fraction of \(C_2H_2\) in the new mixture

Mole fraction \(x_{C_2H_2}=\frac{n_{C_2H_2,\text{remaining}}}{n_{\text{total}}}=\frac{9.4}{27.3}\approx0.344\)

Step8: Calculate mole fraction of \(C_2H_2\) in the initial mixture

Initial total moles \(n_{\text{initial total}}=12.1 + 20.6=32.7\space mol\)
Initial mole fraction \(x_{C_2H_2,\text{initial}}=\frac{12.1}{32.7}\approx0.370\)

Answer:

\(x_{C_2H_2,\text{initial}} = 0.370\); \(x_{C_2H_2,\text{new}} = 0.344\)