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Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . if 15.8 g of sodium sulfate is produced from the reaction of 57.9 g of sulfuric acid and 13.7 g of sodium hydroxide, calculate the percent yield of sodium sulfate. round your answer to 3 significant figures.
Accepted Answer
Here's how to calculate the percent yield of sodium sulfate:
1. Write the balanced chemical equation:
H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)
2. Determine the limiting reactant:
Convert the masses of sulfuric acid and sodium hydroxide to moles using their molar masses (H₂SO₄: 98.08 g/mol, NaOH: 40.00 g/mol):
Moles of H₂SO₄ = 57.9 g / 98.08 g/mol = 0.591 mol
Moles of NaOH = 13.7 g / 40.00 g/mol = 0.343 mol
The stoichiometry of the balanced equation shows that 1 mole of H₂SO₄ reacts with 2 moles of NaOH. Since we have less than twice the moles of NaOH compared to H₂SO₄, NaOH is the limiting reactant.
3. Calculate the theoretical yield of sodium sulfate:
Use the mole ratio from the balanced equation and the moles of the limiting reactant (NaOH) to find the theoretical moles of Na₂SO₄ produced:
Moles of Na₂SO₄ = 0.343 mol NaOH × (1 mol Na₂SO₄ / 2 mol NaOH) = 0.172 mol Na₂SO₄
Convert the moles of Na₂SO₄ to grams using its molar mass (142.04 g/mol):
Theoretical yield of Na₂SO₄ = 0.172 mol × 142.04 g/mol = 24.4 g
4. Calculate the percent yield:
Percent yield = (actual yield / theoretical yield) × 100%
Percent yield = (15.8 g / 24.4 g) × 100% = 64.7%
Therefore, the percent yield of sodium sulfate is 64.7%.
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