How many grams of calcium chloride will be produced when 29.0 g of calcium carbonate is combined with 13.0 g of hydrochloric acid?

19.7 g of CaCl₂ are produced

Explanation:

This is the reaction:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

As we have the mass of reactants, we should determine the moles we used, of each

29g. 1mol / 10.08 g = 0.289 moles

13 g. 1mol / 36.45 g = 0.356 moles

Ratio is 1:2, so If I have 1 mol of carbonate I need the doulbe of hydrochloric. Certainly the limiting reactant is the HCl.

I have 0.289 moles of salt and I need 0.578 (x2) moles of acid. - I only have 0.356 moles.

As we found out the limiting reactant we can work with the reaction. Ratio with CaCl₂ is 2:1 so,

2 moles of HCl produce 1 mol of CaCl₂

Then, 0.356 moles of HCl would produce (0.356. 1) / 2 = 0.178 moles.

To convert the moles to mass we do: 0.178 mol. 110.98 g / 1mol = 19.7 g