Question

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

Answer 1

When 68.1 g of lead (II) sulfate reacts, 22.055 g of sulfuric acid are produced.

Step-by-step explanation:

When lead (II) sulfate reacts, it produces sulfuric acid. In the balanced chemical equation for this reaction:

PbSO4 + H2SO4 → Pb(HSO4)2

We can see that 1 mole of lead (II) sulfate reacts with 1 mole of sulfuric acid. To determine the grams of sulfuric acid produced, we need to convert the grams of lead (II) sulfate to moles using the molar mass, and then use the stoichiometry of the reaction to convert moles of lead (II) sulfate to moles of sulfuric acid, and finally convert moles of sulfuric acid to grams using the molar mass.

Step 1: Convert grams of lead (II) sulfate to moles:

68.1 g PbSO4 × (1 mol PbSO4 / 303.26 g PbSO4) = 0.225 mol PbSO4

Step 2: Use stoichiometry to convert moles of PbSO4 to moles of H2SO4:

0.225 mol PbSO4 × (1 mol H2SO4 / 1 mol PbSO4) = 0.225 mol H2SO4

Step 3: Convert moles of H2SO4 to grams:

0.225 mol H2SO4 × (98.09 g H2SO4 / 1 mol H2SO4) = 22.055 g H2SO4

Answer 2

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