Question

The $\%$ by volume of $C_4{H}_{10}$ in a gaseous mixture of $C_4{H}_{10}$, $C{H}_4$ and CO is $40$. When $200$ ml of the mixture is burnt in excess of $O_2$. Find volume (in ml) of $C{O}_2$ produced.

• A. $220$
• B. $340$
• C. $440$
• D. $560$

Answer 1

The correct option is C $440$

% by volume of $C_4{H}_{10}=40$%

Total volume of mixture=$200mL$

Volume of $C_4{H}_{10}=\frac{40}{100}\times 200$

Volume of $C_4{H}_{10}=80mL$

Remaining Volume=$200-80=120mL$

Volume of ${CH}_4+CO=120mL$

Let Volume of ${CH}_4$ be V mL

then volume of CO will be $120-VmL$

${CH}_4+2O_2⟶{CO}_2+2H_{2}O$

VmL VmL

$CO+\frac{1}{2}{O}_2⟶{CO}_2$

(120-V)mL (120-V)mL

$C_4{H}_{10}+\frac{13}{2}{O}_2⟶4{CO}_2+{5H}_{2}O$

80ml 80x4=320mL

Total volume of ${CO}_2$ produced=$VmL+(120-V)mL+320mL$

$=(V+120-V+320)mL$

$=440mL$