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

Question

How many grams of water are formed when 45 moles of sulfuric acid react with 50 moles of sodium hydroxide? Which reactant is limiting? How many moles of the excess reactant will be left after the reaction?

H₂SO₄ + 2 NaOH   Na₂SO₄ + 2 H₂O

Answer 1

Look at the balanced equation for the reaction:

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

The coefficients tell you the mole ratio. So, 1 mole H₂SO₄ reacts with 2 mol NaOH to => 1mol Na₂SO₄ and 2 mol H₂O.

Since it takes twice as many moles of NaOH, you can tell that NaOH is limiting because you start with 50 mol NaOH and 45 mol H₂SO₄. Clearly not twice as much NaOH as H₂SO₄.

Using dimensional analysis we can find grams of H₂O produced from 50 mols NaOH:

50 mol NaOH x 2 mol H₂O/2 mol NaOH = 50 moles H₂O formed

50 moles H₂O x 18 g/mole H₂O = 900 g H₂O formed

To find moles of excess reactant, again we can use dimensional analysis along with the mole ratios of the balanced equation:

50 mol NaOH x 1 mol H₂SO₄/2 mol NaOH = 25 moles H₂SO₄ used up

moles H₂SO₄ remaining = 45 - 25 = 20 moles H₂SO₄ remaining