When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. CaCO3 ( s ) + 2 HCl ( aq ) ⟶ CaCl2 ( aq ) + H 2 O ( l ) + CO 2 ( g ) How many grams of calcium chloride will be produced when 25.0 g of calcium carbonate is combined with 13.0 g of hydrochloric acid

19.7 g of CaCl₂ are produced in the reaction

Explanation:

In this excersise we need to know the limiting reactant in order to determine the mass produced of a one of the products

The reaction is: CaCO₃ (s) + 2HCl (aq) → CaCl₂(aq) + H₂O (l) + CO₂(g)

First of all we must find the limiting reactant. For that purpose, we convert the mass of reactants to moles

25 g / 100.08 g/mol = 0.249 moles of carbonate

13 g / 36.45 g/mol = 0.357 moles of HCl

We work with the stoichiometry of the reaction:

1 mol of carbonate reacts with 2 moles of hydrochloric

Then, 0.249 moles of carbonate must react with (0.249 . 2) /1 = 0.498 moles of HCl (We do not have enough HCl, so this is the limtiing reactant)

We work with the stoichiometry reactant / product

2 moles of HCl can produce 1 mol of CaCl₂

Therefore 0.357 moles of HCl must produce (0.357 .1) / 2  = 0.178 moles of chloride.

We convert the moles to mass → 0.178 mol . 110.98 g /1mol = 19.7g

where is the following

explanation: