chemistry question

Question

2.  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

Hello, Amanda,

A balanced equation! Thanks. Many reaction questions are not yet balanced. Fun, but tiring.

H₂SO₄ + 2 NaOH --> Na₂SO₄ + 2 H₂O

We want to know how many grams of water we'll get with 60 grams of H₂SO₄ and 200 grams of NaOH. We'll need to convert those masses into moles so that we can compare them properly. Do that by dividing the maas by the molar mass of the compound. I get the follwing:

H₂SO₄ : 60 grams/98 grams/mole = 0.6119 moles H₂SO₄

NaOH: 200 g/40.0 g/mole = 5.001 moles NaOH

The balanced equation says we need twice as many moles of NaOH as we have of H₂SO₄ , a molar ratio of 2/1. We have far more than twice. 2*0.6119 = 1.224 moles NaOH required to consume all of the H₂SO₄ . We'' assume all of the H₂SO₄ is consumed, requiring 1.224 moles of NaOH, leaving behind 3.778 moles of the base. A real waste just to make water, but it is easy. Just don't drink the water with all the unreacted NaOH.

The balanced equation predicts we'll get 2 moles of water for every one mole of H₂SO₄ , a molar ratio of 2/1, again. That means we'll obtain 1.224 moles of water. Multiply by the molar mass of water to find the ggrams of water: (1.224 moles H₂O)*(18g/mole) = 22.03 grams H₂O. We have 2 sig figs, so the result is 22 grams. Plus we'll have 3.778 moles of leftover NaOH, which is a mass of 151 grams. In metric terms, that is a mega-mess. 173 grams of damp sodium hydroxide.

Bob