Reaction Stoichiometry

Question

When 5.32 g Ca react with 14.5 g of HCl according to the equation below:

Ca + 2HCl = CaCl2 + H2

How many liters of hydrogen gas are produced at STP?

(Assuming Ca is the limiting reactant)

Answer 1

Ca(s) + 2HCl(aq) ==> CaCl₂(aq) + H₂(g) ... balanced equation

When given the mass of both reactants, we must first find the limiting reactant.

moles Ca = 5.32 g x 1 mol / 40.1 g = 0.133 mols Ca (÷1->0.133)

moles HCl = 14.5 g x 1 mol / 36.5 g = 0.397 mols HCl (÷2->0.198)

Ca is the limiting reactant (as assumed), so we'll use the moles of Ca (0.133 mols) to find theoretical yield of H2.

0.133 mols Ca x 1 mol H₂ / mol Ca = 0.133 mols H₂

At STP, 1 mol of any ideal gas occupies 22.4 liters

Volume of H₂ = 0.133 mols x 22.4 L / mol = 2.98 L