91Ó°ÊÓ

Given,

Mass of mixture (CH₄ and C₃H₈) = 252g

Mass of CO₂collected = 748 g

Let x be the mass of CH₄ in the mixture and subsequently, 252-x is the mass of C₃H₈ in the mixture. We will convert these mass expressions into the moles of CO₂ produced. Th chemical equations and molar mass needed are the same in part a),

So moles (g) of CO₂ (from CH₄) =

${\mathrm{xgCH}}_4×\frac{1{\mathrm{molCH}}_4}{16.05{\mathrm{gCH}}_4}×\frac{1{\mathrm{molCO}}_2}{1{\mathrm{molCH}}_4}=\frac{\mathrm{x}}{16.05}{\mathrm{molCO}}_2$

Moles(g) of CO₂ (from C₃H₈) =

$\begin{array}{l}(252−\mathrm{x}){\mathrm{gC}}_3{\mathrm{H}}_8×\frac{1{\mathrm{molC}}_3{\mathrm{H}}_8}{44.01{\mathrm{gC}}_3{\mathrm{H}}_8}×\frac{3{\mathrm{molCO}}_2}{1{\mathrm{molC}}_3{\mathrm{H}}_8}\\ =\frac{3(252−\mathrm{x})}{44.11}{\mathrm{molCO}}_2\end{array}$

We have the following expressions,

Moles of CO₂ collected = Moles of CO₂ (from CH₄)+ Moles(g) of CO₂ (from C₃H₈)

$\begin{array}{l}748{\mathrm{gCO}}_2×\frac{1{\mathrm{molCO}}_2}{44.01{\mathrm{gCO}}_2}=\frac{\mathrm{x}}{16.05}{\mathrm{molCO}}_2+\frac{3(252−\mathrm{x})}{44.11}{\mathrm{molCO}}_2\\ \mathrm{x}=25.03\mathrm{g}\end{array}$

We have the following masses of CH₄ and C₃H₈.

Mass of CH₄ = x = 25.03g

Mass of CH₄ = 252-x = 226.97 g

The mass percent of CH₄ in the mixture is as follows:

Mass % of CH₄=$\frac{25.03\mathrm{g}{\mathrm{CH}}_4}{252\mathrm{g}\mathrm{mixture}}×100=9.93\mathrm{\%}$

So the final answer is 9.93% CH₄.