Aqueous sulfuric acid (H2SO4) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (Na2SO4) and liquid water (H2O).

Question

If 3.23g of sodium sulfate is produced from the reaction of 2.9g of sulfuric acid and 3.3g of sodium hydroxide, calculate the percent yield of sodium sulfate. Round your answer to 2 significant figures.

= ______ %

Answer 1

This isn't that much different from your previous question(s). You really should learn how to do these yourself. I don't know if you'll be tested on this, but if you are, you won't do well unless you understand the process. Just my two cents. Now, on to the problem at hand.

As always, you MUST write a correctly balanced equation for the reaction:

H₂SO₄ + 2NaOH ==> Na₂SO₄ + 2H₂O ... balanced equation

Next, find the limiting reactant. I've explained this before in my answers to your previous questions.

2.9 g H₂SO₄ x 1 mol / 98 g = 0.0296 mols H₂SO₄ (÷1->0.0296)

3.3 g NaOH x 1 mol / 40 g = 0.0825 (÷2->0.041)

H₂SO₄ is limiting and will determine the amount of Na₂SO₄ that can form.

Theoretical yield of Na₂SO₄:

0.0296 mols H₂SO₄ x 1 mol Na₂SO₄ / mol H₂SO₄ x 142 g Na₂SO₄ / mol = 4.20 g Na₂SO₄

Percent yield of Na₂SO₄ = actual yield / theoretical yield (x100%)

% yield = 3.23 g / 4.20 g (x100%) = 77% yield (2 sig. figs.)