Chemistry homework

Question

The quantity of atmospheric SO2(g) can be determined by reaction with H2O2(aq) according to the balanced chemical equation shown.

H2O2(aq)+SO2(g)⟶H2SO4(aq)

The amount of sulfuric acid produced can be determined by titration with sodium hydroxide of known concentration. A 753.0 g sample of air that is known to contain sulfur dioxide was reacted with excess hydrogen peroxide. Given that 18.50 mL of 0.00250 M NaOH(aq) was required to neutralize the H2SO4(aq) produced, calculate the mass percentage of

SO2(g) in the air sample.

Answer 1

H₂O₂(aq) + SO₂(g) ==> H₂SO₄(aq) (EQ 1)

H₂SO₄ + 2NaOH ==> Na₂SO₄ + 2H₂O (EQ 2)

From EQ 2, we can find moles H2SO4 that was produced in EQ 1:

moles NaOH needed = 18.50 ml x 1 L/1000 ml x 0.00250 mol/L = 4.625x10^-5 moles NaOH needed

moles H₂SO₄ present = 4.625x10^-5 mol NaOH x 1 mol H₂SO₄ / 2 mol NaOH = 2.31x10^-5 mol H₂SO₄ present

Since from EQ 1 we see that 1 mol H₂SO₄ derives from 1 mol SO₂, we now know that the moles of SO₂ present originally = 2.31x10^-5 mol SO₂

Converting this to grams of SO₂, we have 2.31x10^-5 mol SO₂ x 64.1 g/mol = 1.48x10^-3 g SO₂

Mass percent SO₂ = mass SO₂ / mass of sample (x100) = 1.48x10^-3 g / 753 g (x100) = 1.97x10^-4%