Auseful 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.

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

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consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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