Hale takes notes in class as shown below.
1 answer:
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Answer:
1,812 wt%
Explanation:
The reactions for this titration are:
2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻
I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻
The moles in the end point of S₂O₃⁻ are:
0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:
4,463x10⁻⁴ moles of S₂O₃⁻× = 2,2315x10⁻⁴ moles of I₃⁻
As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:
2,2315x10⁻⁴ moles of I₃⁻× = 4,463x10⁻⁴ moles of Ce(IV). These moles are:
4,463x10⁻⁴ moles of Ce(IV)× = <em>0,0625 g of Ce(IV)</em>
As the sample has a 3,452g, the weight percent is:
0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>
I hope it helps!