A 2.50 g sample of a bronze gong was cut out and dissolved in sulfuric acid. The copper sulfate produced was mixed with KI to form CuI and the triiodide ion, I₃^-.
The I₃^- ion was then titrated with thiosulfate, S₂O₃^2-.
Cu(s) + 2 H₂SO₄(aq) $\rightarrow$ CuSO₄(aq) + SO₂(g) + 2 H₂O(l)
2 CuSO₄(aq) + 5 I^-(aq) $\rightarrow$ 2 CuI(s) + I₃^-(aq) + 2 SO₄^2-(aq)
I₃^-(aq) + 2 S₂O₃^2-(aq) → 3 I^-(aq) + S₄O₆^2-(aq)

If it took 31.5 mL of 1.00 M thiosulfate for this titration, how many moles of copper were present in the original 2.50 g sample?

Question

A 2.50 g sample of a bronze gong was cut out and dissolved in sulfuric acid. The copper sulfate produced was mixed with KI to form CuI and the triiodide ion, I₃^-.
The I₃^- ion was then titrated with thiosulfate, S₂O₃^2-.
Cu(s) + 2 H₂SO₄(aq)

\rightarrow

CuSO₄(aq) + SO₂(g) + 2 H₂O(l)
2 CuSO₄(aq) + 5 I^-(aq)

\rightarrow

2 CuI(s) + I₃^-(aq) + 2 SO₄^2-(aq)
I₃^-(aq) + 2 S₂O₃^2-(aq) → 3 I^-(aq) + S₄O₆^2-(aq)

If it took 31.5 mL of 1.00 M thiosulfate for this titration, how many moles of copper were present in the original 2.50 g sample?

Quizplus · Accepted Answer

The Answer of A 2.50 g sample of a bronze gong was cut out and dissolved in sulfuric acid. The copper sulfate produced was mixed with KI to form CuI and the triiodide ion, I₃^-. The I₃^- ion was then titrated with thiosulfate, S₂O₃^2-. Cu(s) + 2 H₂SO₄(aq) $\rightarrow$ CuSO₄(aq) + SO₂(g)+ 2 H₂O(l) 2 CuSO₄(aq) + 5 I^-(aq) $\rightarrow$ 2 CuI(s) + I₃^-(aq) + 2 SO₄^2-(aq) I₃^-(aq) + 2 S₂O₃^2-(aq) → 3 I^-(aq) + S₄O₆^2-(aq) If it took 31.5 mL of 1.00 M thiosulfate for this titration, how many moles of copper were present in the original 2.50 g sample?

A 250 G Sample of a Bronze Gong Was Cut Out