How many grams of calcium chloride will be produced when 29.0 g of calcium carbonate is combined with 15.0 g of hydrochloric acid?

22.7 g of CaCl₂ are produced in the reaction

Explanation:

This is the reaction:

CaCO₃ +  2HCl  →  CaCl₂  + CO₂  +  H₂O

Now, let's determine the limiting reactant.

Let's divide the mass between the molar mass, to find out moles of each reactant.

29 g / 100.08 g/m = 0.289 of carbonate

15 g / 36.45 g/m = 0.411 of acid

1 mol of carbonate must react with 2 moles of acid

0.289 moles of carbonate will react with the double of moles (0.578)

I only have 0.411 of HCl, so the acid is the limiting reactant.

Ratio is 2:1, so I will produce the half of moles, of salt.

0.411 / 2 = 0.205 moles of CaCl₂

Mol . molar mass = mass → 0.205 m . 110.98 g/m = 22.7 g