A useful application of oxalic acid is the removal of rust (fe2o3) from, say, bathtub rings according to the reaction below. fe2o3 (s) + 6 h2c2o4 (aq) 2 fe (c2o4) 33 - (aq) + 3 h2o + 6 h + (aq) calculate the number of grams of rust that can be removed by 8.00 ✕ 102 ml of a 0.800 m solution of oxalic acid.

Answer: 170.9 g

Explanation:

1) Balanced chemical equation:

Fe₂O₃ (s) + 6 H₂C₂O₄ (aq) → 2Fe (C₂O₄) ₃³⁻ + 3H₂O + 6H⁺ (aq)

2) mole ratio:

1mol Fe₂O₃ : 6 mol H₂C₂O₄

3) number of moles of H₂C₂O₄ in the solution

M = n / V ⇒ n = MV = 0.800M * 0.800 liter = 6.40 moles

4) Porportionality

1mol Fe₂O₃ / 6 mol H₂C₂O₄ = x / 6.40 mol H₂C₂O₄

⇒ x = 1.07 mol Fe₂O₃

5) Convert 1.07 mol Fe₂O₃ to grams.

molar mass Fe₂O₃ = 159.69 g/mol

mass in grams = molaer mass * number of moles

mass in grams = 159.69 g/mol * 1.07 mol = 170.9 g