Answer:
There should react 1008.3 grams KOH
Explanation:
<u>Step 1</u>: Data given
yield = 20.85 %
actual yield of H2O = 67.5 grams
Molar mass of H2SO4 = 98.08 g/mol
Molar mass of H2O = 18.02 g/mol
Molar mass of KOH = 56.11 g/mol
<u>Step 2:</u> The balanced equation
H2SO4 + 2KOH ⇒ K2SO4 + 2H2O
<u>Step 3:</u> Calculate moles of H2O
Moles H2O = Mass H2O / Molar mass H2O
Moles H2O = 67.5 grams / 18.02 g/mol
Moles H2O = 3.746 moles
<u>Step 4:</u> Calculate theoretical yield
% yield= (actual yield/theoretical yield)*100%
Theoretical yield= (Actual yield/Percent yield) * 100%
Theoretical yield = (3.746 moles /20.85) *100%
Theoretical yield = 17.97 moles of H2O
<u>Step 5:</u> Calculate moles of KOH
For 2 moles of H2O produced, we need 2 moles of KOH
For 17.97 moles of H2O produced, we need 17.97 moles of KOH
<u>Step 6:</u> Calculate mass of KOH
Mass KOH =moles KOH * Molar mass KOH
Mass KOH = 17.97 moles * 56.11 g/mol
Mass KOH = 1008.3 grams
There should react 1008.3 grams KOH