Answer:
The theoretical yield of water formed is 2.2 grams
Explanation:
Step 1: Data given
Mass of H2SO4 = 5.9 grams
Mass of NaOH = 6.6 grams
Molar mass H2SO4 = 98.08 g/mol
Molar mass of NaOH = 40.0 g/mol
Step 2: The balanced equation
2NaOH + H2SO4 → Na2SO4 + 2H2O
Step 3: Calculate moles
Moles = mass / molar mass
Moles H2SO4 = 5.9 grams / 98.08 g/mol
Moles H2SO4 = 0.060 moles
Moles NaOH = 6.6 grams / 40.0 g/mol
Moles NaOH = 0.165 moles
Step 4: Calculate the limiting reactant
For 2 moles NaOH we need 1 mol H2SO4 to produce 1 mol Na2SO4 and 2 moles H2O
H2SO4 is the limiting reactant. It will completely be consumed ( 0.060 moles). NaOH is in excess . There will react 2*0.060 = 0.120 moles
There will remain 0.165 - 0.120 = 0.045 moles NaOH
Step 5: Calculate moles H2O
For 2 moles NaOH we need 1 mol H2SO4 to produce 1 mol Na2SO4 and 2 moles H2O
For 0.0600 moles H2SO4 we'll have 2*0.0600 = 0.120 moles H2O
Step 6: Calculate mass H2O
Mass H2O = 0.120 moles * 18.02 g/mol
Mass H2O = 2.16 grams
The theoretical yield of water formed is 2.2 grams
