chemistry question

Question

1. When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. CaCO3(s)+2HCl(aq)⟶CaCl2(aq)+H2O(l)+CO2(g) . How many grams of calcium chloride will be produced when 29.0g of calcium carbonate is combined with 12.0 g of hydrochloric acid? and How many grams of the excess reactant will remain after the reaction is complete?



Answer 1

Molar mass CaCO3 = 100.1 g / mol

Molar mass HCl = 36.5 g / mol

Molar mass CaCl2 = 111 g / mol

(1). Find the limiting reactant

For CaCO3: 29.0 g x 1 mol /100.1 g = 0.2897 mols

For HCl: 12.0 g x 1 mol / 36.5 g = 0.3288 mols

HCl is limiting because according to the balanced equation, we need 2 mols HCl for every 1 mol of CaCO3. We don't have twice as much HCl, so it will run out first.

(2). Use mols of limiting reactant to find amount of product formed:

Grams CaCl2 formed = 0.3288 mol HCl x 1 mol CaCl2 / 2 mol HCl x 111 g CaCl2/mol = 18.2 g CaCl2

(3). To find grams of excess CaCO3 left over, first find moles used, then subtract that from moles present (0.2897) and then convert to grams:

mols CaCO3 used up = 0.3288 mols HCl x 1 mol CaCO3 / 2 mol HCl = 0.1644 mols CaCO3

mols CaCO3 left over = 0.2897 mol - 0.1644 mol = 0.1253 mols left

grams CaCO3 left over = 0.1253 mols x 100.1 g/mol = 12.5 g CaCO3 left over