help with chemistry please?

Question

Calculate the theoretical yield of water in grams that can be made rom the reaction of 250 grams of potassium hydroxide and 350 grams of sulfuric acid. Identify the limiting reagant. Calculate the number of grams of excess reagent remaining after the reaction is completed. If the actual yield of water is 60 grams, calculate the percent yield.

Answer 1

First balance the reaction:

2 KOH + H2SO4 ---> K2SO4 + 2 H2O

250 g KOH / 57 g/mol = 4.39 moles KOH
350 g H2SO4 / 98 g/mol = 3.57 moles H2SO4

Since you don't have twice as many moles of KOH as you do H2SO4, KOH is your limiting reactant. You can get as many moles of H2O as you had KOH to start, or 4.39 moles, which is 79.0 g. 2.195 moles of H2SO4 will react, or 215 g, leaving 135 g unreacted. If you only get 60 g of water, then your percentage yied is 75.9%.