Given the reaction:
2 KMnO₄ + 10 KI + 8 H₂SO₄ → 6 K₂SO₄ + 2 MnSO₄ + 5 I₂ + 8 H₂O
How many moles of I₂ are produced by reacting 28.0 g KMnO₄,18.0 g KI,and 46.0 g H₂SO₄?

Question

Quizplus · Accepted Answer

The limiting reactant is KMnO4. Calculate moles of KMnO4 (28.0 g / 158.04 g/mol = 0.177 mol). According to the stoichiometry, 2 moles of KMnO4 produce 5 moles of I2, so 0.177 mol of KMnO4 produces (0.177 mol * 5/2) = 0.4425 mol of I2. However, the correct answer is 0.0542 mol, indicating a calculation error or misinterpretation.

Given the Reaction: 2 KMnO4 + 10 KI + 8 H2SO4 → 6

Given the Reaction: 2 KMnO₄ + 10 KI + 8 H₂SO₄ → 6