Lead (II) carbonate decomposes to give lead (II) oxide and carbon dioxide: PbCO 3 (s) PbO (s) CO 2 (g) grams of lead (II) oxide will be produced by the decomposition of 8.75 g of lead (II) carbonate

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