Answer to Question #340198 in Chemistry for Andrew

Question

When a solution containing excess Sodium sulfate is added to a second solution containing 3.18 g of Barium nitrate, Barium sulfate precipitation. a. Calculate the theoretical yield of barium sulfate. b. If the actual yield is 2.69 g. Calculate the percent yield.

Answer 1

Solution:

The molar mass of barium nitrate, Ba(NO₃)₂, is 261.337 g/mol

Therefore,

Moles of Ba(NO₃)₂ = [3.18 g Ba(NO₃)₂] × [1 mol Ba(NO₃)₂ / 261.337 g Ba(NO₃)₂] = 0.01217 mol

Balanced chemical equation:

Na₂SO₄(aq) + Ba(NO₃)₂(aq) → BaSO₄(s) + 2NaNO₃(aq)

According to stoichiometry:

1 mol of Ba(NO₃)₂ produces 1 mol of BaSO₄

Thus, 0.01217 mol of Ba(NO₃)₂ produce:

[0.01217 mol Ba(NO₃)₂] × [1 mol BaSO₄ / 1 mol Ba(NO₃)₂] = 0.01217 mol BaSO₄

The molar mass of barium sulfate, BaSO₄, is 233.38 g/mol

Therefore,

Mass of BaSO₄ = [0.01217 mol BaSO₄] × [233.38 g BaSO₄ / 1 mol BaSO₄] = 2.84 g BaSO₄

The theoretical yield of barium sulfate (BaSO₄) is 2.84 grams

Percent yield = (Actual yield / Theoretical yield) × 100%

Percent yield = (2.69 g / 2.84 g) × 100% = 94.72%

Percent yield = 94.72%

Answers:

(a): The theoretical yield of barium sulfate (BaSO₄) is 2.84 grams

(b): The percent yield is 94.72%