Potassium permanganate reacts with sulfuric acid as per the equation shown below. Indicate the amount of excess reactant after the reaction of 7.24g KMnO4 and 10.32g H2SO4? 4KMnO4+6H2SO4 = 2K2SO4 +4MnSO4 +6H2O + 5O2

Calculate moles KMnO4 and convert to moles of the product.

Calculate moles H2SO4 and convert to moles of the product.
The smaller number of moles of the product is the correct value and the material producing that value is the limiting reagent.

To determine the amount of excess reactant, we need to find the limiting reactant first.

Step 1: Calculate the molar masses of each reactant:
- Molar mass of KMnO4: 39.1 g/mol (K) + 54.9 g/mol (Mn) + 64.0 g/mol (O4) ≈ 158.0 g/mol
- Molar mass of H2SO4: 2.0 g/mol (H) + 32.1 g/mol (S) + 4.0 g/mol (O4) ≈ 98.1 g/mol

Step 2: Convert the masses of the reactants to moles:
- Moles of KMnO4 = mass / molar mass = 7.24 g / 158.0 g/mol ≈ 0.0460 mol
- Moles of H2SO4 = mass / molar mass = 10.32 g / 98.1 g/mol ≈ 0.105 mol

Step 3: Use the balanced equation to determine the stoichiometric ratio between the reactants. From the balanced equation, we can see that the ratio between KMnO4 and H2SO4 is 4:6.

Step 4: Compare the moles of each reactant based on the stoichiometric ratio. Since the ratio is 4:6, we need to multiply the moles of KMnO4 by (6/4) to compare them correctly.

- Moles of KMnO4 required = 0.0460 mol × (6/4) = 0.069 mol

Step 5: Determine the limiting reactant.

Comparing the moles of H2SO4 (0.105 mol) with the calculated moles of KMnO4 required (0.069 mol), it is clear that KMnO4 is the limiting reactant. This means that H2SO4 is in excess.

Step 6: Calculate the amount of excess reactant.

To find the excess reactant, subtract the moles of the limiting reactant from the total moles of the excess reactant.

- Moles of H2SO4 in excess = Moles of H2SO4 - Moles of H2SO4 required
- Moles of H2SO4 in excess = 0.105 mol - 0.069 mol = 0.036 mol

Step 7: Convert the excess moles of H2SO4 back to grams by multiplying by the molar mass:

- Mass of H2SO4 in excess = Moles of H2SO4 in excess × molar mass of H2SO4
- Mass of H2SO4 in excess = 0.036 mol × 98.1 g/mol ≈ 3.53 g

Therefore, the amount of excess reactant (H2SO4) after the reaction of 7.24 g KMnO4 and 10.32 g H2SO4 is approximately 3.53 grams.