A Hydrocarbon Problem

C₃H₈ + 3O₂ -> 3CO₂ + 4H₂, CH₄ + O₂ -> CO₂ + 2H₂, Now we assume x moles of CH₄ and y moles of C₃H₈ in the original mixture. So seeing above two questions we can conclude that the total CO₂ liberated is (x+3y)moles (satisfying stoichiometry!). Now for ideal gases P V=n R*T. And we are given that volume and temperature are same so we can say P∝n. Therefore (x+y)/(x+3y)=320/448. From here we get y=64. So mole Fraction of C₃H₈ is y/(x+y). 64/320 = 0.2!

P V = n R T for unchanged V and T with constant R:

$\frac{P_1}{n_1} = \frac{P_2}{n_2}$

$\frac{320}{n_1} = \frac{448}{n_2}$

$\frac{n_2}{n_1} = \frac{7}{5}$

$CH_4 + 2 O_2 \rightarrow CO_2 + 2 H_2O$

$C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O$

Pressures given only concern with $CH_4, C_3H_8~and~CO_2,$ where $H_2O$ could be liquid while $O_2$ is not inclusive in figures mentioned. By trial and error starting from $CH_4 : C_3H_8 = 1 : 1$ but 2.5 + 2.5 for 2. 5 + 7.5 = 10,

4 portions of $CH_4$ shall give 4 portions of $CO_2$

1 portion of $C_3H_8$ shall give 3 portions of $CO_2$

Consequently, 5 portions for 7 portions is resulted by having $\frac{1}{1 + 4}$ of $C_3H_8.$

Answer: $\boxed{0.2}$