Question

Oxalic acid can remove rust (Fe2O3) caused by bathtub rings according to the reaction Fe2O3(s) - 6H2C2O4(aq) rightarrow 2Fe(C2O4)33-(aq) - 3H2O(l) - 6H+(aq) C calculate the number of grams of rust that can be removed by 6.00 times 102 mL of a 0.100 M solution of oxalic acid.

Answer 1

Answer: The mass of rust that can be removed is 1.597 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$     .....(1)

Molarity of oxalic acid solution = 0.1255 M

Volume of solution = $6.00\times 10^2mL$ = 600 mL = 0.600 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of oxalic acid}}{0.600L}\\\\\text{Moles of oxalic acid}=(0.100mol/L\times 0.600L)=0.06mol$

For the given chemical reaction:

$Fe_2O_3(s)+6H_2C_2O_4(aq.)\rightarrow 2Fe(C_2O_4)_3^{3-}(aq.)+3H_2O(l)+6H^+(aq.)$

By Stoichiometry of the reaction:

6 moles of oxalic acid reacts with 1 mole of ferric oxide (rust)

So, 0.06 moles of oxalic acid will react with = $\frac{1}{6}\times 0.06=0.01mol$ of ferric oxide (rust)

To calculate the mass of rust for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of rust (ferric oxide) = 159.7 g/mol

Moles of rust = 0.01 moles

Putting values in above equation, we get:

$0.01mol=\frac{\text{Mass of rust}}{159.7g/mol}\\\\\text{Mass of rust}=(0.01mol\times 159.7g/mol)=1.597g$

Hence, the mass of rust that can be removed is 1.597 grams