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Standard XII

Chemistry

Stoichiometric Calculations

The percentag...

Question

The percentage by volume of $C_{3} H_{8}$ in a mixture of $C_{3} H_{8}, C H_{4}$ and $C O$ is $36.5$ . The volume of $C O_{2}$ produced (in mL), when $100$ mL of the mixture is burnt in excess of $O_{2}$ , is :

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Solution

Let the mixture contains $a, b,$ and $c$ mL of $C_{3} H_{8}, C H_{4}$ and $C O$ respectively, then
$a + b + c = 100 . . . (i)$
Also, $a = 36.5 . . . (i i)$
The combustion reactions are:
$C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
$C O + \frac{1}{2} O_{2} \to {C O}_{2}$
$∴ C O_{2}$ formed as a combustion of mixture
$= 3 a + b + c (s e e v o l u m e o r m o l e r a t i o)$
$= 3 \times 36.5 + (b + c)$
$= 3 \times 36.5 + (100 - 36.5) = 173 m L$

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The percentage by volume of C3H8 in a mixture of C3H8, CH4 and CO is 36.5. The volume of CO2 produced (in mL), when 100 mL of the mixture is burnt in excess of O2, is :

The percentage by volume of $C_{3} H_{8}$ in a mixture of $C_{3} H_{8}, C H_{4}$ and $C O$ is $36.5$ . The volume of $C O_{2}$ produced (in mL), when $100$ mL of the mixture is burnt in excess of $O_{2}$ , is :