Answer:
We will have 7.30 grams lead(II) oxide
Explanation:
Step 1: Data given
Mass of lead (II)carbonate = 8.75 grams
Molar mass PbCO3 = 267.21 g/mol
Step 2: The balanced equation
PbCO3 (s) ⇆ PbO(s) + CO2(g)
Step 3: Calculate moles PbCO3
Moles PbCO3 = mass / molar mass
Moles PbCO3 = 8.75 grams / 267.21 g/mol
Moles PbCO3 = 0.0327 moles
Step 4: Calculate moles PbO
For 1 mol PbCO3 we'll have 1 mol PbO and 1 mol CO2
For 0.0327 moles PbCO3 we'll have 0.0327 moles PbO
Step 5: Calculate mass PbO
Mass PbO = moles PbO * molar mass PbO
Mass PbO = 0.0327 moles * 223.2 g/mol
Mass PbO = 7.30 grams
We will have 7.30 grams lead(II) oxide
