Question

Sulfur is an undesirable impurity in coal and petroleum fuels. The mass percentage in fuel can be determined by burning the fuel in oxygen and dissolving the resulting SO3 gas in water to form sulfuric acid. In one experiment, 8.54 g of fuel was burned, and the resulting sulfuric acid was tritiated to an endpoint with 17.54 mL of 0.100 M NaOH (aq).

A.) determine the amount (in moles) of the sulfuric acid that was produced

B.) what is the mass percentage of sulfur in the fuel?

Answer 1

(A.) The balanced reaction is : H₂SO₄ + 2 NaOH Na₂SO₄ + 2 H₂O

concentration NaOH = 0.100 M

volume NaOH = 17.54 mL = 0.01754 L

moles NaOH required = (concentration NaOH) * (volume NaOH in Liter)

moles NaOH required = (0.100 M) * (0.01754 L)

moles NaOH required = 0.001754 mol

moles H₂SO₄ produced = (moles NaOH required) * (1 mole H₂SO₄ / 2 moles NaOH)

moles H₂SO₄ produced = (0.001754 mol) * (1/2)

moles H₂SO₄ produced = (0.001754 mol) * (0.5)

moles H₂SO₄ produced = 8.77 x 10^-4 mol

(b.) moles sulfur in fuel = moles H₂SO₄ produced

moles sulfur in fuel = 8.77 x 10^-4 mol

mass sulfur in fuel = (moles sulfur in fuel) * (molar mass sulfur)

mass sulfur in fuel = (8.77 x 10^-4 mol) * (32.065 g/mol)

mass sulfur in fuel = 0.0281 g

mass percentage of sulfur in the fuel = (mass sulfur in fuel / mass of fuel) * 100

mass percentage of sulfur in the fuel = (0.0281 g / 8.54 g) * 100

mass percentage of sulfur in the fuel = (0.00329) * 100

mass percentage of sulfur in the fuel = 0.329 %