Chemical Reactions

Question

75.0 grams of aqueous sulfuric acid was reacted with 99.0 g grams of aqueous sodium hydroxide to create aqueous sodium sulfate and liquid water. How many grams of sodium sulfate would be formed?

Answer 1

Any time you are given the amounts of both reactants, it will be a limiting reactant problem. One easy way to find the limiting reactant is to divide the moles of each reactant by the corresponding coefficient in the balanced equation. The lower value indicates the limiting reactant.

H₂SO₄(aq) + 2NaOH(aq) ==> Na₂SO₄(aq) + 2H₂O(l) .. balanced equation

moles H₂SO₄ = 75.0 g x 1 mol / 98 g = 0.765 moles H₂SO₄(÷1->0.77)

moles NaOH = 99.0 g x 1 mol / 40 g = 2.48 moles NaOH (÷2->1.24)

In this case, H₂SO₄ is limiting as 0.77 is less than 1.24

Use the MOLES of the limiting reactant (0.765 moles) to find out how much product is formed:

0.765 mols H₂SO₄ x 1 mol Na₂SO₄ / mol H₂SO₄ x 142 g / mol = 109 g Na₂SO₄ formed