Question

How many milliliters of 0.204 Mol KMnO4 are needed to react with 3.24 g of iron(II) sulfate, FeSO4? The reation is as folows. 10FeSO4(aq) + 2 KMnO4(aq) = 5Fe2(SO4)3(aq) + 2MnSO4(aq) + K2SO4(aq) + 8H2O(l)​

Answer 1

Step-by-step explanation:

nFeSo4=3.36/152

nkmno4=1/5nFeSO4

V=17.68 ml

Answer 2

21.0 mL of 0.204 M KMnO4 are needed to react with 3.24 g of FeSO4.

Here's how to solve the problem:

Step 1: Calculate the moles of FeSO4

1. Molar mass of FeSO4 = 55.8 (Fe) + 32.07 (S) + 4 * 16 (O) = 151.91 g/mol

2. Moles of FeSO4 = 3.24 g / 151.91 g/mol = 0.0214 mol

Step 2: Use the balanced equation to find the moles of KMnO4

From the balanced equation, 10 moles of FeSO4 react with 2 moles of KMnO4:

10 FeSO4 + 2 KMnO4 → 5 Fe2(SO4)3 + 2 MnSO4 + K2SO4 + 8 H2O

Therefore, the moles of KMnO4 needed is:

0.0214 mol FeSO4 * (2 mol KMnO4 / 10 mol FeSO4) = 0.00428 mol KMnO4

Step 3: Calculate the volume of KMnO4 solution

1. Molarity of KMnO4 solution = 0.204 mol/L

2. Volume of KMnO4 solution = moles of KMnO4 / molarity of KMnO4 solution

3. Volume of KMnO4 solution = 0.00428 mol / 0.204 mol/L = 0.0210 L = 21.0 mL

Therefore, 21.0 mL of 0.204 Mol KMnO4 are needed to react with 3.24 g of iron(II) sulfate.