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Stoichiometric Calculations

69 g of K 2 C...

Question

69 g of $K_{2} C O_{3}$ is thermally decomposed to give $K_{2} O$ and $C O_{2}$ (g). Calculate the volume of $C O_{2}$ formed at STP. (Molar mass of $K_{2} C O_{3} = 138 g m o l^{- 1})$

22.4 L

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11.2 L

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33.6 L

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5.6 L

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Solution

The correct option is B 11.2 L
$K_{2} C O_{3} (s) \to K_{2} O (s) + C O_{2} (g)$

$Applying the principle of atom conservation (POAC) to C, Moles of C in K_{2} C O_{3} = moles of carbon in C O_{2} 1 \times Number of mol of K_{2} C O_{3} = 1 \times number of moles of C O_{2}$
Hence, the number of moles of $C O_{2}$ produced will be equal to the number of moles of $K_{2} C O_{3} decomposed.$
Now we have,
$Moles of K_{2} C O_{3} = \frac{G i v e n w e i g h t}{M o l a r M a s s} = \frac{69}{138} = 0.5 m o l$
$Moles of K_{2} C O_{3} = Moles of C O_{2} = 0.5 m o l We know that the volume of 1 mol of gas at STP is 22.4 L and volume of C O_{2} = number of mol of C O_{2} \times 22.4 L$ = 0.50 × 22.4 = $11.2$ L
$Therefore, the volume of C O_{2} = 11.2 L$

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