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

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?

Quizplus · Accepted Answer

First, we need to convert the given masses into moles using their molar masses:
- KMnO4: 28.0 g / 158.03 g/mol = 0.177 mol
- KI: 18.0 g / 166.00 g/mol = 0.108 mol
- H2SO4: 46.0 g / 98.08 g/mol = 0.469 mol

Now, we can use the stoichiometry of the reaction to determine the limiting reagent (the reactant that will be completely consumed) and the amount of I2 produced:

- From the balanced equation, we can see that 2 moles of KMnO4 react with 10 moles of KI to produce 5 moles of I2.
- The mole ratio of KMnO4 to KI is 2:10 = 1:5, which means that the limiting reagent is KMnO4 because we have less of it compared to what is needed to react with all of the KI.
- Using the mole ratio of KMnO4 to I2, we can calculate how many moles of I2 will be produced:
0.177 mol KMnO4 x (5 mol I2 / 2 mol KMnO4) = 0.443 mol I2

Therefore, the answer is B) 0.0542 mol.

Given the Reaction: 2 KMnO4 + 10 KI + 8

Given the Reaction: 2 KMnO₄ + 10 KI + 8