Question

Aqueous sulfuric acid H2SO4 will react with solid sodium hydroxide NaOH to produce aqueous sodium sulfate Na2SO4 and liquid water H2O

50. g of sulfuric acid is mixed with 15.1 g of sodium hydroxide. Calculate the minimum mass of sulfuric acid that could be left over by the chemical reaction. Round your answer to 2

significant digits.

Answer 1

The balanced equatiton is

H2SO4 + 2 NaOH -----> Na2SO4 + H2O

number of moles of H2SO4 = 50.0g / 98.079 g/mol = 0.5097 mole

number of moles of NaOH = 15.1 g / 40.0 g/mol = 0.3775 mole

from the balanced equation we can say that

1 mole of H2SO4 requires 2 mole of NaOH so

0.5097 mole of H2SO4 will require 1.019 mole of NaOH

but we have 0.3775 mole of NaOH so NaOH will be the limiting reactant and H2SO4 is an excess reactant

2 mole of NaOH requires 1 mole of H2SO4 so

0.3775 mole of NaOH will require 0.1887 mole of H2SO4

Therefore, the number of moles of H2SO4 left oveer = 0.5097 - 0.1887 = 0.321 mole of H2So4

1mole of H2SO4 =98.097 g

so 0.321 mole of H2SO4 = 31.5 g

Therefore, the mass of H2SO4 left over = 31.5 g