A 40.0mL of a mixture of $H_{2}and{O}_2$ was placed in a gas burette at 18°C and 1atm. Pressure. The mixture when subjected to an electric spark formed water.  The remaining pure gas had a volume of 10mL at 18°C and 1atm pressure.  If the remaining gas is $H_2$, which of the following is/are correct 

No. of moles of ${\text{H}}_{\text{2}}{\text{\hspace{0.17em}\&\hspace{0.17em}O}}_{\text{2}}$ in 40 ml $\text{=\hspace{0.17em}}\frac{\text{1\hspace{0.17em}×\hspace{0.17em}40}}{\text{1000\hspace{0.17em}×\hspace{0.17em}0.0821\hspace{0.17em}×\hspace{0.17em}291}}{\text{\hspace{0.17em}=\hspace{0.17em}1.67\hspace{0.17em}×\hspace{0.17em}10}}^{\text{-3}}$

 No. of moles of ${\text{H}}_2$ left $=\frac{1\times 10}{1000\times 0.0821\times 291}=4.18\times {10}^{-4}$

Total number of moles of ${\text{H}}_{\text{2}}{\text{\hspace{0.17em}and\hspace{0.17em}O}}_{\text{2}}$ react in the mole ratio of 2 : 1

$\therefore$ No. of moles of out of $1.252\times {10}^{-3}=\frac{1-252\times {10}^{-3}{x}^2}{3}=8.35\times {10}^{-4}$

Total no of moles of $H_2=8.35\times {10}^{-4}+4.18\times {10}^{-4}=12.53\times {10}^{-4}$

% of $H_2$ in the original mixture = $\frac{12.53\times {10}^{-4}\times 100}{1.67\times {10}^{-3}}=75.0$