determining limiting reactants etc

Question

One method for manufacturing sodium sulfate proceeds according to the following equation:

4NaCl + 2SO2

+ 2H2O + O2  2Na2SO4 + 4HCl

Address the following questions about this reaction (4 each)

a. If 2335 g of sodium chloride are available to react with 1355 g of sulfur dioxide, determine

the limiting reagent.

b. Determine the mass of oxygen required for the complete consumption of the limiting reagent.

c. Determine the mass of sodium sulfate produced if the limiting reagent is completely

consumed and the percent yield is 82.5%.

Answer 1

a) (2335g NaCl) * (1 mole NaCl /58.44g NaCl) * (2 mole Na₂SO₄ /4 mole NaCl) = 19.98 mole Na₂SO₄.

(1355g SO₂) * (1 mole SO₂ /64.06g SO₂) * (2 mole Na₂SO₄ /2 mole SO₂) = 21.15 mole Na₂SO₄.

Na is limiting, with some SO₂ in excess.

b) (19.98 mole Na₂SO₄) * (1 mole O₂ /2 mole Na₂SO₄) * (32.00g O₂ /1 mole O₂) = 319.6 g O₂.

c) (19.98 mole Na₂SO₄) * (142.04 g Na₂SO₄ /1 mole Na₂SO₄) = 2838g Na₂SO₄ theoretical yield.

Percent yield = 100 actual/theoretical, so above result * 0.825 = 2341g Na₂SO₄ actual yield.