Question in the description, can you please explain how you got the answer, thank you

Question

How many grams of water can be produced with 60 grams of sulfuric acid and 200 grams of sodium hydroxide? Which reactant is limiting? How many grams of the excess reactant will be left after the reaction?

H₂SO₄ + 2 NaOH   Na₂SO₄ + 2 H₂O

Answer 1

H₂SO₄ + 2 NaOH ==> Na₂SO₄ + 2 H₂O ... Balanced equation

60 g..........200 g.............................? g..........

Finding limiting reactant:

60 g H₂SO₄ x 1 mol H₂SO₄/98 g = 0.612 mol H₂SO₄ x 2 mol H₂O/mol H₂SO₄ x 18 g H₂O/mol = 22 g H₂O

200 g NaOH x 1 mol NaOH/40 g = 5 mol NaOH x 2 mol H₂O / 2 mol NaOH x 18 g H₂O/mol = 180 g H₂O

H₂SO₄ is the limiting reactant.

Grams excess reactant (NaOH) left over will be initial amount present - amount used.

Initial NaOH present = 200 g

Amount NaOH used up = 0.612 mol H₂SO₄ x 2 mol NaOH/mol H₂SO₄ x 40 g NaOH/mol = 49 g

Grams NaOH left over = 200 g - 49 g = 151 g NaOH left over