What is the Answer to the Following?

Question

KOH with a mass of 50.0 grams is neutralized by 20.0 g of sulfuric acid. The products are potassium sulfate and water

A. How many moles of potassium sulfate produced

B. What is the grams of water produced

Answer 1

A) 0.204 moles.

The balanced reaction is 2 KOH + H2SO4 => K2SO4 + 2 H2O.

First off, this is a limiting reaction--if you calculate how much sulfuric acid is needed to neutralize the KOH, you will get 43.2 g:

(50 g KOH) * (1 mol KOH / 56.11g KOH) * (1mol H2SO4 / 2mol KOH) * (98.08g H2SO4 / 1 mol H2SO4) = 43.2 g.

Thus, the H2SO4 is the limiting reagent.

From here:

(20 g H2SO4) * (1 mol H2SO4 / 98.08g H2SO4) * (1 mol K2SO4 / 1 mol H2SO4) = 0.204 moles K2SO4.

B) Continuing from the results of A:

(0.204 moles K2SO4) * (2 mol H2O / 1 mol K2SO4) * (18.02 g H2O / 1 mol H2O) = 7.35 grams H2O.