question practice chemistry

Question

1.      What is the greatest amount of sodium chloride in moles that can be made with 13 grams copper II chloride and 15 grams 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

     How many grams of water are formed when 45 moles of sulfuric acid react with 50 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

1) (13g CuCl₂) * (1 mol CuCl₂ / 134.4g CuCl₂) * (2 mol NaCl / 1 mol CuCl₂) = 0.19₃ mol NaCl.

(15g NaNO₃) * (1 mol NaNO₃ /85.0g NaNO₃) * (2 mol NaCl / 2 mol NaNO₃) = 0.17₆ mol NaCl.

Sodium nitrate is limiting, with 0.01₇ mole excess of copper chloride (this result has only 1 sig fig, 0.02 mole, but the answer would be wrong if you rounded the individual mole numbers before taking the difference).

2) (45 mol H₂SO₄) * (2 mol H₂O / 1 mol H₂SO₄) = 90 mol H₂O.

(50 mol NaOH) * (2 mol H₂O / 2 mol NaOH) = 50 mol H₂O.

Sodium hydroxide is limiting, with 40 moles excess of sulfuric acid.