Question

An useful application of oxalic acid is the removal of rust $\left(F{e}_2{O}_3\right)$ from a cloth.
$F{e}_{2}C{O}_3\left(s\right)+6H_2{C}_2{O}_4\to 2F{e}_2{(C_2{O}_4{)}_3}^{3-}+3H_{2}O+H^{+}$
The number of grams (as nearest integer) of rust that can be removed by $5.00\times {10}^2$ mL of $1$ M oxalic acid is :

Answer 1

The balanced chemical equation is as follows:
$F{e}_2{O}_3\left(s\right)+6H_2{C}_2{O}_4\to 2Fe(C_2{O}_4{)}_3^{-3}+3H_{2}O+6H^{+}$
Thus, $1$ mole of rust requires $6$ moles of oxalic acid.
The number of moles of oxalic acid is $\frac{5.00\times {10}^2\left(ml\right)}{1000}\times 1\left(M\right)=0.5$ moles.
The number of moles of rust $=\frac{0.5}{6}=0.0833$ mol.
The mass of the rust that can be removed $=0.0833\times 159.69=13.33$ g.
The nearest integer is $13$.