Lets say the reaction H2SO4 + 2KOH - > K2SO4 + 2H2O, is known to proceed by a 20.85% yield via a particular procedure.

How many grams of KOH must be reacted with an excess of H2SO4, in order to collect 67.5g of H2O?

There should react 1008.3 grams KOH

Explanation:

Step 1: 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

Step 2: The balanced equation

H2SO4 + 2KOH ⇒ K2SO4 + 2H2O

Step 3: 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

Step 4: 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

Step 5: 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

Step 6: 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