Question

121 grams of iron (II) acetate reacts with 360 grams of lead (II) oxalate. Predict the products, write the complete reaction equation, balance, and calculate the mass of the precipitate.

Answer 1

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