Question 3.3.4: Calcium carbonate reacts with hydrochloric acid to form calc......
Calcium carbonate reacts with hydrochloric acid to form calcium chloride, carbon dioxide, and water.
Unbalanced equation: CaCO_{3} + HCl → CaCl_{2} + CO_{2} + H_{2}O
In one reaction, 54.6 g of CO_{2} is produced. What amount (in mol) of HCl was consumed? What mass (in grams) of calcium chloride is produced?
You are asked to calculate the amount (in mol) of a reactant consumed and the mass (in grams) of a product formed in a reaction.
You are given the unbalanced equation and the mass of one of the products of the reaction.
Step 1. Write a balanced chemical equation.
CaCO_{3} + 2 HCl → CaCl_{2} + CO_{2} + H_{2}O
Step 2. Calculate moles of CO_{2} produced.
54.6 g CO_{2} × \frac{1\text{ mol CO}_{2}}{44.01\text{ g}} = 1.24 mol CO_{2}
Step 3. Calculate moles of HCl consumed and moles of CaCl_{2} produced using stoichiometric factors derived from the balanced chemical equation.
1.24 mol CO_{2} × \frac{2\text{ mol HCl}}{1\text{ mol CO}_{2}} = 2.48 mol HCl
1.24 mol CO_{2} × \frac{1\text{ mol CaCl}_{2}}{1\text{ mol CO}_{2}} = 1.24 mol CaCl_{2}
Step 4. Calculate mass (in grams) of CaCl_{2} produced.
1.24 mol CaCl_{2} × \frac{111.0\text{ g}}{1\text{ mol CaCl}_{2}} = 138 g CaCl_{2}
Notice that it is possible to set up both calculations as a single step.
54.6 g CO_{2}\left(\frac{1\text{ mol CO}_{2}}{44.01\text{ g}} \right) \left(\frac{2\text{ mol HCl}}{1\text{ mol CO}_{2}} \right) = 2.48 mol HCL
54.6 g CO_{2}\left(\frac{1\text{ mol CO}_{2}}{44.01\text{ g}} \right)\left(\frac{1\text{ mol CaCl}_{2}}{1\text{ mol CO}_{2}} \right)\left(\frac{111.0\text{ g}}{1\text{ mol CaCl}_{2}}\right) = 138 g CaCl_{2}
Is your answer reasonable? More than one mole of CO_{2} is produced in the reaction, so the amounts of HCl consumed and CaCl_{2} produced are greater than one mole.