Question

Aqueous sulfuric acid H₂SO₄ reacts with solid sodium hydroxide NaOH to produce aqueous sodium sulfate Na₂SO₄ and liquid water H₂O. If 6.41gof sodium sulfate is produced from the reaction of 7.8g of sulfuric acid and 4.4g of sodium hydroxide, calculate the percent yield of sodium sulfate.

Be sure your answer has the correct number of significant digits in it.

Answer 1

Reaction between H₂SO₄ and NaOH

H₂SO₄ + 2NaOH Na₂SO₄ + 2H₂O

molar mass of H₂SO₄ = 98.079 gm/mol that mean 1 mole of H₂SO₄ = 98.079 gm

molar mass of NaOH = 39.997 gm/mol that mean 2 mole of NaOH = 79.994 gm

According to reaction 1 mole of H₂SO₄ react with 2 mole of NaOH that mean 98.079 gm of H₂SO₄ react with 79.994 gm of NaOH then 7.8 gm of H₂SO₄ require 7.8 79.994/98.079 = 6.36 gm of NaOH

but NaOH given only 4.4 gm thus NaOH is limiting reactant react completly.

molar mass of Na₂SO₄ = 142.04 gm/mol that mean 1 mole of Na₂SO₄ = 142.04 gm

According to reaction 2 mole of NaOH produce 1 mole of Na₂SO₄ that mean 79.994 gm of NaOH produce 142.04 gm of Na₂SO₄ then 4.4 gm of NaOH produce 4.4 142.04/79.994 = 7.8127859 gm of Na₂SO₄

7.8127859 gm is therotical yield

7.8127859 gm = 100% then 7.8 gm = 7.8 100/ 7.8127859 = 99.8363 %

percent yield = 99.8363%