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Question

Calcium carbonate reacts with aqueous HCl 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

1000 ml of 0.75 M Hd have 0.75 mol of Hd=(0.75)(36.5)=24.375 g
mass of Hd in 25 ml of 0.75 M Hd=24.3751000×25 g=0.6844 g
CaCO3(5)+2HCl(aq)CaCl2(aq)+CO2(g)H2O(l)
2 moles of Hd i.e., 73 g HCl react completely with 1 mole of CaCO3 i.e., 100 grams
0.6844 Hd reacts with CaCO3=(10073)(0.6844)=0.938 g

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