Question

Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. 5.90 g of sulfuric acid and 5.90 g of lead(II) acetate are mixed. Calculate the number of grams of sulfuric acid present in the mixture after the reaction is complete.

Answer 1

The mass of the sulfuric acid that is left when the reaction is complete is 4.1 g.

Explain the reaction

We have the reaction equation given as;

`Pb(CH3COO)_2 + H_2SO_4 ---- > PbSO_4 + 2C_2H_4O_2`

Number of moles of sulfuric acid = 5.90 g/98 g/mol

= 0.06 moles

Number of moles of lead(II) acetate = 5.90 g/325 g/mol

= 0.018 moles

If 1 mole of lead(II) acetate reacts with 1 mole of sulfuric acid then the reaction is 1:1.

Lead(II) acetate is the limiting reactant

Number of moles of sulfuric acid left over = 0.06 moles - 0.018 moles

= 0.042 moles

Mass of left over sulfuric acid = 0.042 moles * 98 g/mol

=4.1 g