Answer to Question #340202 in Chemistry for Andrew

Question #340202

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.

1
Expert's answer
2022-05-16T04:17:05-0400

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%

The percent yield is 94.72%


Answers:

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

(b): The percent yield is 94.72%

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