Using the balanced equation below, how many grams of lead (II) sulfate would be produced from the complete reaction of 23.6 g lead (IV) oxide?

The chemical equation is:

Pb + PbO2 + 2 H2SO4 → 2 PbSO4 + 2 H2O

From the periodic table:

mass of Pb = 207.2 grams

mass of oxygen = 16 grams

mass of sulfur = 32 grams

Therefore:

molar mass of PbO2 = 207.2 + 2 (16) = 239.2 grams

molar mass of PbSO4 = 207.2 + 32 + 4 (16) = 303.2 grams

From the balanced equation, one mole of PbO2 produces two moles of PbSO4. This means that 239.2 grams of PbO2 produces 606.4 grams of PbSO4.

Use cross multiplication to find the amount of PbSO4 produced from 23.6 grams of PbO2 as follows:

amount = (23.6*606.2) / (239.2) = 59.8287 grams