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Stoichiometric Calculations

The minimum a...

Question

The minimum amount of $A g_{2} C O_{3} (s)$ required to produce sufficient oxygen for the complete combustion of $C_{2} H_{2}$ , which produces $11.2 L$ of $C O_{2}$ at STP after combustion is: $[A g = 108]$
$A g_{2} C O_{3} (s) \to 2 A g (s) + C O_{2} (g) + \frac{1}{2} O_{2} (g)$
$C_{2} H_{2} + \frac{5}{2} O_{2} \to 2 C O_{2} + H_{2} O$

276 g

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345 g

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690 g

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1380 g

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Solution

The correct option is B $345 g$
The balanced reaction given for the complete combustion of $C_{2} H_{2}$ is
$C_{2} H_{2} + \frac{5}{2} O_{2} \to 2 C O_{2} + H_{2} O$

So to produce $11.2 L$ of $C O_{2}$ , $C_{2} H_{2}$ required is $= 5.6 L = \frac{5.6}{22.4} m o l = 0.25 m o l$

Again, for complete combustion of $1 m o l o f C_{2} H_{2}$ , $\frac{5}{2} m o l O_{2} r e q u i r e d$

$∴$ For 0.25 mol of $C_{2} H_{2}$ , $\frac{5}{2} \times 0.25 m o l = 0.625 m o l o f O_{2}$ required.

The balanced chemical equation for the production of $O_{2}$ from $A g_{2} C O_{3}$ is,
$A g_{2} C O_{3} (s) \to 2 A g (s) + C O_{2} (g) + \frac{1}{2} O_{2} (g)$

So for the production of $0.625 m o l O_{2}$ minimum amount of $A g_{2} C O_{3} r e q u i r e d = 0.625 \times 2 m o l = 1.25 m o l = 1.25 \times 276 = 345 g$

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