Chemistry Stoichiometry

Question

If 10.00 grams of CaCO₃ reacts with 15.0 grams of HCl, how many grams of CO₂ is produced?

CaCO₃ (s) +  2 HCl (aq) ⟶ CaCl₂ (aq) + CO₂ (g) + H₂O (l) 

Answer 1

CaCO₃ (s) +  2 HCl (aq) ⟶ CaCl₂ (aq) + CO₂ (g) + H₂O (l)  .. balanced equation

Use the stoichiometry (mole ratios) from the balanced equation, along with dimensional analysis to find the answer to the question. Note: since we are given the amounts of BOTH reactants, we must FIRST find which reactant is present in limiting supply, and then use that to determine how much CO₂ is produced.

One way to find the limiting reactant is to simply divide the mols of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less represents the limiting reactant.

molar mass CaCO₃ = 100.1 g / mol

10.00 g CaCO₃ x 1 mol / 100.1 g = 0.1000 mols CaCO₃ (÷1->0.100)

molar mass HCl = 36.5 g / mol

15.0 g HCl x 1 mol HCl / 36.5 g = 0.411 mols HCl (÷2->0.205)

Since 0.100 is less than 0.205, CaCO₃ is the limiting reactant. Now use the moles of CaCO₃ (0.100) to calculate the mass of CO₂ produced, as follows:

molar mass CO₂ = 44.0 g / mol

0.100 mols CaCO₃ x 1 mol CO₂ / mol CaCO₃ x 44.0 g CO₂ / mol = 4.4 g CO₂