85.2 grams of sodium sulfate and 66.7 grams of lead (II) chloride react. Write the balanced equation. Classify the reaction. Identify the precipitate Identify the limiting and excess reactant.

Question

 85.2 grams of sodium sulfate and 66.7 grams of lead (II) chloride react.

1. Write the balanced equation
2. Name each of the products
3. Classify the reaction
4. Identify the precipitate
5. Identify the limiting and excess reactant
6. How much of each product is made?
7. What is the actual yield of the solid product obtained during the experiment if the percent yield is 75%

Answer 1

a. Na₂SO₄ (aq) + PbCl₂ (aq) → 2 NaCl (aq) + PbSO₄ (s)

b. Products formed are sodium chloride and lead (II) sulfate.

c. Reaction is a double displacement.

d. Precipitate formed is lead (II) sulfate because it is insoluble in water.

e. Convert g of Na₂SO₄ to g of PbCl₂. Then compare how many grams of PbCl₂ are needed to react with 85.2 g of Na₂SO₄. (NOTE! You can convert g of PbCl₂ to g of Na₂SO₄ and perform a similar analysis).

85.2 g Na₂SO₄ x (1 mol Na₂SO₄ / 142.04 g Na₂SO₄) x (1 mol PbCl₂ / 1 mol Na₂SO₄) x (278.10 g PbCl₂ / 1 mol PbCl₂) = 167 g PbCl₂

As you can see, 167 g of PbCl₂ are needed to completely react with 85.2 g of Na₂SO₄. Since only 66.7 g of PbCl₂ are used, Na₂SO₄ is therefore, the excess reagent and PbCl₂ is the limiting reagent.

f. 66.7 g PbCl₂ x (1 mol PbCl₂ / 278.10 g PbCl₂) x (2 mol NaCl / 1 mol PbCl₂) x (58.44 g NaCl / 1 mol NaCl) = 28.0 g NaCl

66.7 g PbCl₂ x (1 mol PbCl₂ / 278.10 g PbCl₂) x (1 mol PbSO₄ / 1 mol PbCl₂) x (303.26 g PbSO₄ / 1 mol PbSO₄) = 72.7 g PbSO₄

g. 72.7 g PbSO₄ = Theoretical yield.

%-Yield = (Actual yield / Theoretical yield) x 100%

75% = Actual yield / 72.7 g

Actual yield = 75% x 72.7 g = 54.5 g