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Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction given below:

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

What mass of CaCl2 will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of CaCO3? Name the limiting reagent. Calculate the number of moles of CaCl2 formed in the reaction.

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According to given equation 1 mol of CaCO3 (s) requires 2 mol of HCl (aq). Hence, for the reaction of 10 mol of CaCO3 (s) number of moles of HCl required would be:

But we have only 0.19 mol HCl (aq), hence, HCl (aq) is limiting reagent. So amount of CaCl2 formed will depend on the amount of HCl available. Since, 2 mol HCl (aq) forms 1 mol of CaCl2, therefore, 0.19 mol of HCl (aq) would give:

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