How many grams of sulfuric acid are produced when 68.1 g of lead (II) sulfate react?

The chemical reaction to produce sulfuric acid fro lead (ii) sulfate is:

PbSO4 + SO2 + 2H2O - - > Pb + 2H2SO4

first convert the mass of PbSO4 to moles

moles PbSO4 = 68.1 g (1 mol / 303.26 g)

moles PbSO4 = 0.2246 mol PbSO4

mass H2SO4 = 0.2246 mol PbSO4 (2 mol H2SO4 / 1 mol PbSO4) (98 g / mol)

mas H2SO4 = 44 g H2SO4