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Stoichiometric Calculations

A sample of c...

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A sample of coal gas contains $50 %$ $H_{2}$ , $30 %$ $C H_{4}$ , $14 %$ $C O$ and $6 %$ $C_{2} H_{4}$ . $100$ mL of coal gas is mixed with $150$ mL of $O_{2}$ and the mixture is exploded. The volume of $C O_{2}$ which will be formed (in ml) is :

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Solution

$H_{2} + \frac{1}{2} O_{2} \to H_{2} O$
$50$ mL
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
$30$ mL
$C O + \frac{1}{2} O_{2} \to C O_{2}$
$14$ mL
$C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O$
$6$ mL
Volume of $O_{2}$ required $= \frac{50}{2} + 30 \times 2 + \frac{14}{2} + 6 \times 3$ $=$ $25 + 60 + 7 + 18$ $=$ $110$ mL.
$∴$ Volume of $O_{2}$ left $=$ 1 $50 - 110$ $=$ $40$ mL.
Volume of $C O_{2}$ formed $=$ $30$ + $14$ + $6 \times 2$
$=$ $56$ mL

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