A mixture of $CH X_{4} (g)$ and $H X_{2} (g)$ containing $0.50$ mol in total is burnt completely (to $CO X_{2} (g)$ and $H X_{2} O (l)$ ) and releases $300 kJ$ of heat. The standard enthalpies of combustion are $- 890 kJ mo l^{- 1}$ for $CH X_{4}$ and $- 286 kJ mo l^{- 1}$ for $H X_{2}$ . The mole fraction of $CH X_{4}$ in the mixture is

Answer: A

Let

y

mol be

CH X_{4}

890 y + 286 (0.50 - y) = 300

, so

604 y = 300 - 143 = 157

and

y = 0.260 mol

. Mole fraction of

CH X_{4}

= y /0.50 = 0.52

. The value

0.48

is the mole fraction of

H X_{2}

0.34

300/890

(ignores the hydrogen), and

0.26

is the number of moles of

CH X_{4}

, not its mole fraction.