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A 1.50 g sample of sodium nitrite, NaNO₂, is dissolved and made up to 250 mL…

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A $1.50\ \text{g}$ sample of sodium nitrite, $\text{NaNO}_2$ , is dissolved and made up to $250\ \text{mL}$ . A $25.0\ \text{mL}$ portion is acidified and needs $32.0\ \text{mL}$ of $0.0250\ \text{mol/L}$ potassium permanganate: $2\text{MnO}_4^- + 5\text{NO}_2^- + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{NO}_3^- + 3\text{H}_2\text{O}$ . The molar mass of $\text{NaNO}_2$ is $69.0\ \text{g/mol}$ . Find the permanganate used, the nitrite in the portion and the percentage of $\text{NaNO}_2$ in the sample.

Need a reminder? The key ideaShow

Moles = mass in g divided by molar mass in g/mol, or concentration in mol/L times volume in litres. Take every mole ratio from the balanced equation, and follow the chain one step at a time.

The worked solution

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← A solution holds 0.0600 mol of potassium manganate…Hot, acidified potassium permanganate oxidises toluene to benzoic acid: 5C₆H₅CH₃ + 6MnO₄⁻ + 18H⁺ →… →

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