Answer:
Mass H2SO4 = 3.42 grams
Mass of lead acetate = 0 grams
Mass PbSO4 = 4.58 grams
Mass of CH3COOH = 1.81 grams
Explanation:
Step 1: Data given
Mass of sulfuric acid = 4.90 grams
Molar mass of sulfuric acid = 98.08 g/mol
Mass of lead acetate = 4.90 grams
Molar mass of lead acetate = 325.29 g/mol
Step 2: The balanced equation
H2SO4 + Pb(C2H3O2)2 → PbSO4 + 2CH3COOH
Step 3: Calculate moles
Moles = mass / molar mass
Moles H2SO4 = 4.90 grams / 98.08 g/mol
Moles H2SO4 = 0.0500 moles
Moles lead acetate = 4.9 grams / 325.29 g/mol
Moles lead acetate = 0.0151 moles
Step 4: Calculate the limiting reactant
For 1 mol H2SO4 we need 1 mol lead acetate to produce 1 mol PbSO4 and 2 moles CH3COOH
The limiting reactant is lead acetate. It will completzly be consumed (0.0151 moles). H2SO4 is in excess. There will react 0.0151 moles. There will remain 0.0500 - 0.0151 = 0.0349 moles
Step 5: Calculate moles of products
For 1 mol H2SO4 we need 1 mol lead acetate to produce 1 mol PbSO4 and 2 moles CH3COOH
For 0.0151 moles lead acetate we'll have 0.0151 moles PbSO4 and 2*0.0151 = 0.0302 moles CH3COOH
Step 6: Calculate mass
Mass = moles * molar mass
Mass H2SO4 = 0.0349 moles * 98.08 g/mol
Mass H2SO4 = 3.42 grams
Mass PbSO4 = 0.0151 moles * 303.26 g/mol
Mass PbSO4 = 4.58 grams
Mass of CH3COOH = 0.0302 moles * 60.05 g/mol
Mass of CH3COOH = 1.81 grams
