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

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"