A white solid is a mixture of sodium ethaneiodate (Na2C2O4), ethanoic acid dihydrate (H2C2O4·2H2O) and an insol - AQA - A-Level Chemistry - Question 11 - 2017 - Paper 1

From the first titration reaction, the stoichiometry shows that 1 mole of MnO4⁻ reacts with 5 moles of C2O4²⁻:

Thus, the moles of sodium ethaneiodate (C2O4²⁻) in the first titration:

$\text{Moles of C2O4²⁻} = 5 \times 5.30 \times 10^{-4} = 2.65 \times 10^{-3} \text{ mol}$

The second titration uses the equation:

$H2C2O4 + 2NaOH \rightarrow Na2C2O4 + 2H2O$

From the second titration, the moles of NaOH used:

$\text{Moles of NaOH} = 0.100 \times 0.01045 = 1.045 \times 10^{-3} \text{ mol}$

Since 2 moles of NaOH react with 1 mole of H2C2O4:

$\text{Moles of H2C2O4} = \frac{1.045 \times 10^{-3}}{2} = 5.225 \times 10^{-4} \text{ mol}$

The moles of sodium ethaneiodate produced:

$\text{Moles of Na2C2O4} = 5.225 \times 10^{-4} \text{ mol}$

Total moles of sodium ethaneiodate in the original sample is:

$\text{Total moles of C2O4²⁻} = 2.65 \times 10^{-3} + 5.225 \times 10^{-4} = 3.173 \times 10^{-3} \text{ mol}$

Using the molar mass of sodium ethaneiodate (Na2C2O4 = 166.02 g/mol), the mass can be determined:

$\text{Mass} = \text{Moles} \times \text{Molar mass} = 3.173 \times 10^{-3} \times 166.02 = 0.526 g$

The percentage by mass of sodium ethaneiodate in the white solid:

$\text{Percentage} = \left( \frac{0.526}{1.90} \right) \times 100 = 27.68\%$

Thus, rounding to the appropriate number of significant figures gives approximately 27.7%.