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Stoichiometric Calculations

If 20 g of ...

Question

If 20 g of $C a C O_{3}$ is treated with 20 g of HCl, how many grams of $C O_{2}$ can be generated according to the following equation ?
$C a C O_{3} + 2 H C l \to C a C l_{2} + H_{2} O + C O_{2}$
(Molecular masses; $C a C O_{3} = 100 u, H C l = 36.5 u C O_{2} = 44 u)$

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Solution

$C a C O_{3} (s) 1 m o l 100 g + 2 H C l (a q .) 2 m o l 73 g \to C a C l_{2} (a q .) + H_{2} O (l) + C O_{2} (g) 1 m o l 44 g$
Let $C a C O_{3} (s)$ be completely consumed in the reaction.
$∵ 100 g C a C O_{3}$ give $44 g C O_{2}$
$∴ 20 g C a C O_{3}$ will give $\frac{44}{100} \times 20 g C O_{2} = 8.8 g C O_{2}$
Let $H C l$ be completely consumed.
$∵$ 73 g $H C l$ give 44 g $C O_{2}$
$∴$ 20 g $H C l$ will give $\frac{44}{73} \times 20 g C O_{2} = 12.054 g C O_{2}$
Since, $C a C O_{3}$ gives least amount of product $C O_{2}$ , hence $C a C O_{3}$ is limiting reactant. Amount of $C O_{2}$ formed will 8.8 g.

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