Question

The total pressure or a mixture of oxygen and hydrogen is $1.0atm$. The mixture is ignited and the water is removed. The remaining gas is pure hydrogen and exerts a pressure of $0.40atm$ when measured at the same values of $T$ and $V$ as the original mixture. What was the composition of the original mixture in mole per cent?

• A. $x_{O_2}=0.2;x_{H_2}=0.8$
• B. $x_{O_2}=0.4;x_{H_2}=0.6$
• C. $x_{O_2}=0.6;x_{H_2}=0.4$
• D. $x_{O_2}=0.8;x_{H_2}=\mathrm{0.2.}$

Answer 1

The correct option is A $x_{O_2}=0.2;x_{H_2}=0.8$

${2H}_2+O_2⟶2H_{2}O$

We know $P\alpha n$

$\therefore$ total moles $=3$

i.e. it contains $\frac{2}{3}{H}_2$ and $\frac{1}{3}{O}_2$

after igniting,

$P=1-0.4=0.6$ was consumed

$\therefore$ $\frac{2\times 0.6}{3}$ mol of $H_2$ and $\frac{0.6}{3}$ mol of $O_2$

$\therefore$ the initial pressure $O_2=0.2$ mole percent

$H_2=0.4+0.4=0.8$