Answer:
The mass of the precipitate is 100.2 grams
Explanation:
Step 1: Data given
Mass of iron (II) acetate = 121.0 grams
Mass of lead (II) oxalate = 360.0 grams
Molar mass Fe(C2H3O2)2 = 173.93 g/mol
Molar mass PbC2O4 = 295.22 g/mol
Step 2: The balanced equation
Fe(C2H3O2)2 + PbC2O4 → FeC2O4 + Pb(C2H3O2)2
Step 3: Calculate moles iron (II) acetate
Moles = mass / molar mass
Moles = 121.0 grams / 173.93 g/mol
Moles = 0.696 moles
Step 4: Calculate moles lead (II) oxalate
Moles = 360.0 grams / 295.22 g/mol
Moles = 1.219 moles
Step 5: Calculate limiting reactant
For 1 mol iron (II) acetate we need 1 mol lead (II) oxalate to produce 1 mol Iron(II) oxalate and 1 mol Lead(II) acetate
iron (II) acetate is the limiting reactant. It will completely be consumed ( 0.696 moles). Lead (II) oxalate is in excess. There will react 0.696 moles. There will remain 1.219 - 0.696 = 0.523 moles
Step 6: Calculate moles of iron(II) oxalate
For 1 mol iron (II) acetate we need 1 mol lead (II) oxalate to produce 1 mol Iron(II) oxalate and 1 mol Lead(II) acetate
For 0.696 moles iron(II) acetate we'll have 0.696 moles iron(II) oxalate
Step 7: Calculate mass iron(II) oxalate
Mass = moles * molar mass
Mass = 0.696 moles * 143.91 g/mol
Mass = 100.2 grams
The mass of the precipitate is 100.2 grams
