Question

Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. 5.60 g of sulfuric acid and 5.60 g of lead(II) acetate are mixed.

PartA: Calculate the number of grams of sulfuric acid present in the mixture after the reaction is complete.
PartB: Calculate the number of grams of lead(II) acetate present in the mixture after the reaction is complete.
PartC: Calculate the number of grams of lead(II) sulfate present in the mixture after the reaction is complete.
PartD: Calculate the number of grams of acetic acid present in the mixture after the reaction is complete

Answer 1

a) The mass of the sulfuric acid is 3.92 g

b) The mass of the lead II acetate is 0 g

c) The mass of the lead II sulfate is 5.2 g

d) The mass of acetic acid is 2.04 g.

The reaction equation is;

`Pb(CH_3COO)_2 + H_2SO_4 --- > PbSO_4 + 2C_2H_4O_2`

Moles of sulfuric acid = 5.60 g /98 g/mol

= 0.057 moles

Moles of lead(II) acetate = 5.60 g /325 g/mol

= 0.017 moles

If the reaction is 1:1

Moles of sulfuric acid left = 0.057 moles - 0.017 moles

= 0.04 moles

Mass of sulfuric acid left = 0.04 moles * 98 g/mol

= 3.92 g

Mass of lead(II) acetate present in the mixture after the reaction is complete = 0 because this is the limiting reactant

Mass of lead II sulfate obtained = 0.017 moles * 303 g/mol

= 5.2 g

If 1 mole of lead(II) acetate produces 2 moles of acetic acid

0.017 moles of lead(II) acetate produces 0.017 moles * 2 moles/1 mole

= 0.034 moles

Mass of the acetic acid = 0.034 moles * 60 g/mol

= 2.04 g