Chemistry Forum

Iron(II) can be oxidized to iron(III) by dichromate ion, which is reduced to chromium(III) in acid solution.
A 2.5000-g sample of iron ore is dissolved and the iron converted into iron(II). Exactly 19.17 mL of 0.0100 M \(Na_2Cr_2O_7\) is required in the titration. What percentage of the ore sample was iron?

What are your own ideas

1. Develop the redox reaction
2. Calculate the moles of dichromate used during titration
3. How many moles iron correspond to it according reaction equation
4. Convert the moles to mass
5. Calculate the percentage with given mass of ore.

The balanced chemical equation for the reaction is:

\( 6 Fe^⁺² + Cr_2O_7^^-2 + 14 H^+ \rightarrow 6 Fe^⁺³ + 2 Cr^⁺³ + 7 H_2O \)

From the equation, we can see that 6 moles of \(Fe^²⁺\) react with 1 mole of \(Cr_2O_7^^2-\). Therefore, the number of moles of \(Fe^²⁺\) in the sample is:

\(n(Fe^²⁺) = \frac{0.0100 mol}{L} \times 19.17 mL \times \frac{1 L}{1000 mL} \times \frac61 = 0.0011542\) mol

The molar mass of Fe is 55.845 g/mol, so the mass of iron in the sample is:

\(m(Fe) = n(Fe^²⁺) \times M(Fe) = 0.0011542 mol \times 55.845 g/mol = 0.06452 g\)

The percentage of iron in the sample is:

\(\% iron = m(Fe) \div m(sample) \times 100\% = 0.06452 g \div 2.5000 g \times 100\% = 2.58\%\)

Therefore, the ore sample is \(2.58\%\) iron.

Note: I don't understand why doesn't Latex work here?

Note: I don't understand why doesn't Latex work here?

I think you have some Syntax error in.

\(\% iron = m(Fe) \div m(sample) \times 100\% = 0.06452 g \div 2.5000 g \times 100\% = 2.58\%\)