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:
Explanation:
nFeSo4=3.36/152
nkmno4=1/5nFeSO4
V=17.68 ml