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Question

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

CaCO3(s)+2HCl(aq)CaCl2(aq)+CO2(g)+H2O(l).

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

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Solution

The mass of HCl is 0.75×36.51000×25=0.684g.

1 mole of calcium carbonate reacts with 2 moles of HCl.

Hence, the mass of calcium carbonate that will react completely with 0.684 g of HCl is

1002×36.5×0.684=0.937g.

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