Question

0.700 g of an impure fecl₃ sample are dissolved in water to create 50.0 ml of solution. it took 1.54 x 10⁻³ mol h₂c₂o₄ to reach the equivalence point.
2fe³⁺(aq) + h₂c₂o₄(aq) → 2co₂(g) + 2h⁺(aq) + 2fe²⁺(aq)
how many moles of fecl₃ react?
? × 10⁻? mol fecl₃

Explanation:

Step1: Identify the mole ratio from the balanced equation.

From the reaction \(2\text{Fe}^{3+}(\text{aq}) + \text{H}_2\text{C}_2\text{O}_4(\text{aq})
ightarrow 2\text{CO}_2(\text{g}) + 2\text{H}^+(\text{aq}) + 2\text{Fe}^{2+}(\text{aq})\), the mole ratio of \(\text{Fe}^{3+}\) (from \(\text{FeCl}_3\)) to \(\text{H}_2\text{C}_2\text{O}_4\) is \(2:1\). Since each \(\text{FeCl}_3\) provides one \(\text{Fe}^{3+}\), the mole ratio of \(\text{FeCl}_3\) to \(\text{H}_2\text{C}_2\text{O}_4\) is also \(2:1\).

Step2: Calculate moles of \(\text{FeCl}_3\) using the mole ratio.

Given moles of \(\text{H}_2\text{C}_2\text{O}_4 = 1.54\times 10^{-3}\) mol. Let moles of \(\text{FeCl}_3\) be \(n\). Using the ratio \(\frac{n}{\text{moles of } \text{H}_2\text{C}_2\text{O}_4}=\frac{2}{1}\), we get \(n = 2\times1.54\times 10^{-3}\) mol.
Calculating \(2\times1.54\times 10^{-3}=3.08\times 10^{-3}\) mol.

Answer:

\(3.08\times 10^{-3}\)