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1.35 Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction,

$CaCO_3 (s) + 2 HCl (aq) \rightarrow CaCl_2 (aq) + CO_2 (g) + H_2 O(l)$

What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?

Answers (1)

best_answer

$0.75M$ $HCl$ contains $0.75\ mole$ in $1000mL$ of solution.

Or, $0.75\times36.5g = 24.375g$ of $HCl$ in $1000mL$ solution.

Therefore,

Mass of $HCl$ in $25mL$ of $0.75M\ HCl$ :

$= \frac{24.375}{1000}\times25 g = 0.6844g$

so, from the given chemical equation,

$CaCO_3 (s) + 2 HCl (aq) \rightarrow CaCl_2 (aq) + CO_2 (g) + H_2 O(l)$

1 mole of $CaCO_{3}(s)$ i.e., $100g$ reacts with 2 moles of $HCl(aq)$ i.e., $73g$ .

Therefore, $0.6844g$ $HCl$ reacts completely with $CaCO_{3}$ to give:

$=\frac{100}{73}\times 0.6844g = 0.938 g$

Posted by

Divya Prakash Singh

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