Question

A solution of sodium hydroxide was neutralized by sulfuric acid. The volume of sodium hydroxide solution used was 675.0 mL, and the concentration was 0.875 M. Calculate the moles of sulfuric acid that were neutralized (assuming that the reaction goes to completion). Give your answer to three significant figures. 2NaOH (aq) + H2SO4 (aq) ? Na2SO4 (aq) + 2H2O (l

Answer 1

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

A solution of sodium hydroxide was neutralized by sulfuric acid. The volume of sodium hydroxide solution used was 940.0 mL, and the concentration was 0.225 M. Calculate the moles of sulfuric acid that were neutralized (assuming that the reaction goes to completion). Give your answer to three significant figures.

2NaOH (aq) + H₂SO₄ (aq) --> Na₂SO₄ (aq) + 2H₂O (l)

2NaOH (aq) + H₂SO₄ (aq) --> Na₂SO₄ (aq) + 2H₂O (l)

from the reaction

2 moles of NaOH required 1 mole of H2S04

moles of NaOH = molarity x volume / 1000

moles of NaOH = 0.225 x 940 / 1000

moles of NaOH = 0.2115

2 moles of NaOH requires 1 mole of H2S04

let 0.2115 moles of NaOH requires y moles of H2S04

y = 0.2115 x 1 /2

y = 0.10575

y = 0.106

so 0.106 moles of H2S04 is neutralized