Question

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 1

To calculate the grams of rust that can be removed by oxalic acid, we use stoichiometry. Given 8.00 ✕ 10^2 ml of 0.800 M oxalic acid, we calculate the moles of oxalic acid, then use the ratio between oxalic acid and rust to find the moles of rust. Finally, we convert moles of rust to grams.

Step-by-step explanation:

To 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, we need to first determine the stoichiometry of the reaction between oxalic acid and rust. From the balanced equation:

Fe2O3(s) + 6 H2C2O4(aq) -> 2 Fe(C2O4)33-(aq) + 3 H2O + 6 H+(aq)

We can see that for every 6 moles of H2C2O4 (oxalic acid), 2 moles of Fe2O3 (rust) are consumed. We can use this ratio to calculate the moles of rust removed by oxalic acid.

Given that we have 8.00 ✕ 102 ml (0.8 L) of a 0.800 M solution of oxalic acid, we can calculate the moles of oxalic acid:

Moles of H2C2O4 = molarity x volume = 0.800 M x 0.8 L = 0.640 moles

Since the ratio between H2C2O4 and Fe2O3 is 6:2, we can use this ratio to determine the moles of Fe2O3:

Moles of Fe2O3 = (6/2) x 0.640 = 1.920 moles

Finally, we can convert moles of Fe2O3 to grams using the molar mass of Fe2O3 (159.69 g/mol):

Grams of Fe2O3 = 1.920 moles x 159.69 g/mol = 306.88 grams

Answer 2

Answer: 170.9 g

Step-by-step 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