Q.A $1.00\ \text{g}$ sample of an iron ore, containing iron entirely as $\text{Fe}^{2+}$, is dissolved and titrated against $0.0167\ \text{M}\ \text{K}_2\text{Cr}_2\text{O}_7$ solution, requiring $25.0\ \text{mL}$ for complete oxidation. Using $\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^{+} \to 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}$ and $M(\text{Fe}) = 56\ \text{g mol}^{-1}$, calculate the percentage of iron in the ore.

[!TLDR] Moles of dichromate from molarity/volume, scaled 1:6 to Fe²⁺, converted to mass of Fe, then to a percentage of the 1.00 g …

Moles of $\text{K}_2\text{Cr}_2\text{O}_7$ used: $0.0167\ \text{mol L}^{-1} \times 0.0250\ \text{L} = 4.175\times10^{-4}\ \text{mol}$. From $\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^{+} \to 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}$, the mole ratio of $\text{Cr}_2\text{O}_7^{2-}$ to $\text{Fe}^{2+}$ is $1:6$, so moles of $\text{Fe}^{2+} = 6 \times 4.175\times10^{-4} = 2.505\times10^{-3}\ \text{mol}$. Mass of iron $= 2.505\times10^{-3}\ \text{mol} \tim …

Compute moles of dichromate from its molarity and volume, scale by the 1:6 mole ratio to find moles of Fe²⁺, convert to mass using the molar mas …

Using a 1:1 ratio instead of 1:6 between dichromate and Fe²⁺; forgetting to convert th …