practice #3 questions

Question

1.      What is the greatest amount of copper IV nitrate in moles that can be made with 24.3 mol copper II chloride and 20.9 mol sodium nitrate? Which reactant is limiting? How many moles of the excess reactant will be left after the reaction? CuCl₂  + 2 NaNO₃ ----->Cu(NO₃)₂ + 2 NaCl

2.   How many moles of sodium sulfate are formed when 13.2 moles of sulfuric acid reacts with 30.6 moles of sodium hydroxide? Which reactant is limiting? How many moles of the excess reactant will be left after the reaction? H₂SO₄ + 2 NaOH  --> Na₂SO₄ + 2 H₂O

Answer 1

CuCl₂  + 2 NaNO₃ ----->Cu(NO₃)₂ + 2 NaCl ... balanced equation

Find limiting reactant by dividing mols of each by corresponding coefficient in balanced equation:

For CuCl₂: 24.3 mols / 1 = 24.3

For NaNO₃: 20.9 / 2 = 10.45

LIMITING REACTANT = NaNO₃

Maximum mols of Cu(NO₃)₂ = 20.9 mols NaNO₃ x 1 mol Cu(NO₃)₂ / 2 mols NaNO₃ = 10.5 mols Cu(NO₃)₂

Mols excess (CuCl₂) left over will be initial mols less mols used up.

mols CuCl₂ used up = 20.9 mols NaNO₃ x 1 mol CuCl₂ / mol NaNO₃ = 20.9 mols CuCl₂ used up

mols left over = 24.3 mols - 20.9 mols = 3.4 mols CuCl₂ left over