Question

One of the uses of oxalic acid, (H2C2O4) is rust removal. It reacts with rust, (Fe2O3) according to the equation:

2Fe2O3(s) + 6H2C2O4(aq) = 2Fe2(C2O4)3^-3(aq) + 3H2O(l) + 6H+(aq)

Calculate the number of grams of rust that can be removed by 5x10^2mL of a 0.100 mol L-1 solution of oxalic acid.

Answer 1

As per the balanced chemical equation (given equation is balanced!).

2 moles of Fe2O3 requires 6 moles of oxalic acid to completely react.

Let's first find moles of oxalic acid,

we have 5*10^2 mL of 0.100M solution = (5*10^2 mL)*(0.100 /1000 mol/mL) = 0.05 moles

the 0.05mol of oxalic will react with = (2*0.05)/6 = 0.0167 moles of Fe2O3

0.0167 moles of Fe2O3 = 0.0167mol * 159.69 g/mol = 2.667 g of Fe2O3.

Thus  5x10^2mL of a 0.100 mol L-1 solution of oxalic acid will be able to remove, 2.667 g of Fe2O3 (rust).

Please note that I have not used the sig. fig. in calculations as it is not specifically asked for. If you are looking for the answer in particular sig. fig. please let me know, I will update accordingly.