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

Question

How many moles of sodium sulfate are formed when 13.2 moles of sulfuric acid reacts with 30.6 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

Hi Aaron,

For this problem, you will want to use the mole ratio for each reactant and then find the amount of sodium sulfate formed. First, make sure the equation is balanced by making sure there are the same number of each element on each side of the equation. This looks good, so we can move onto the mole ratio stoichiometry.

13.2 mol H₂SO₄ x (1 mol Na₂SO₄ / 1 mol H₂SO₄) = 13.2 mol Na₂SO₄

30.6 mol NaOH x (1 mol Na₂SO₄ / 2 mol NaOH) = 15.3 mol Na₂SO₄

You will end up with the lower end amount of Na₂SO₄ because you know that reagent will run out first. Therefore, your limiting reactant is H₂SO₄ and you will create 13.2 mol of Na₂SO₄

To figure out how much NaOH will be left over, you will want to work backward using the amount of Na₂SO₄ you know will be formed.

13.2 mol Na₂SO₄ x (2 mol NaOH / 1 mol Na₂SO₄) = 26.4 mol NaOH

You will need to use 26.4 mol of NaOH, which means you will have 4.2 mol NaOH left over at the end.