This balanced equation shows the reaction of sodium hydroxide and sulfuric acid: 2naoh + h2so4 → na2so4 + 2h2o. in a laboratory experiment, a student mixes 355 grams of sulfuric acid with an excess of sodium hydroxide. what is the theoretical mass of sodium sulfate produced? refer to the periodic table and the polyatomic ion resource.

514.53 g.

Explanation:

The balanced equation of the reaction is:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.

It is clear that 2.0 moles of NaOH reacts with 1.0 mole of H₂SO₄ to produce 1.0 mole of Na₂SO₄ and 2.0 moles of H₂O.The limiting reactant is NaOH and H₂SO₄ is an unlimited reactant.We can calculate the no. of moles of H₂SO₄ using the data available:

n = mass / molar mass = (355.0 g) / (98.0 g/mol) = 3.62 mol.

Using cross multiplication; we can calculate the no. of moles of sodium sulfate Na₂SO₄ can the reaction produce:

1.0 mole of H₂SO₄ produces → 1.0 mole of Na₂SO₄, from the stichiometry.

3.62 moles of H₂SO₄ produces → ??? mole of Na₂SO₄.

The no. of moles of  sodium sulfate Na₂SO₄ can the reaction produce = (1.0 mol)(3.62 mol) / (1.0 mol) = 3.62 mol.Now, we can calculate the grams (mass) of sodium sulfate Na₂SO₄ can the reaction produce = n x molar mass = (3.62 mol)(142.04 g/mol) = 514.53 g.