Question

Aqueous sulfuric acid H2SO4 will react with solid sodium hydroxide NaOH to produce aqueous sodium sulfate Na2SO4 and liquid water H2O . Suppose 81. g of sulfuric acid is mixed with 36.5 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits.

Answer 1

First we pass both reactants to moles:

81 grams of H₂SO₄ * (1 mol / 98.079 grams) = 0.82586 moles of H₂SO₄

36.5 grams of NaOH * (1 mol / 40 grams) = 0.9125 moles of NaOH

Then we express the reaction:

2NaOH + H2SO4 -> 2H2O + Na2SO4

We now determine which is the limiting reactant. In this case, the limiting reactant is sodium hydroxide, as it will run out first.

We convert the moles of NaOH to moles of water:

0.9125 moles of NaOH * (2 moles of H₂O / 2 moles of NaOH) = 0.9125 moles of H₂O

We convert these moles to grams:

0.9125 moles of H₂O * (18 grams / mol) = 16.425 grams

To three significant digits, 16.4 grams.