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Calcium carbonate reacts with aqueous HCl to give CaCl 2 and CO 2 according to the reaction, CaCO 3 (s) + 2 HCl (aq) gives CaCl 2 (aq) + CO 2 (g) + H 2 O(l) What mass of CaCO 3 is required to react completely with 25 mL of 0.75 M HCl?

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)
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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

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