Potassium permanganate(KMnO4) reacts with oxalic acid (H2C2O4) in aqueous sulfuric acid according to: 2KMnO4 5H2C2O4 3H2SO4->2MnSO4 10 CO2 * H2O K2SO4 How many milliliters of a .250 M KMnO4 solution are needed to react completely with 3.225 G of oxalic acid

57.3mL of 0.250M KMnO4 are required

Explanation:

3.225g of oxalic acid are:

3.225g×(1mol / 90.03g) = 0.03582mol of oxalic acid

Based on the reaction, 5 moles of oxalic acid react with 2 moles of KMnO4, thus, for a complete reaction of oxalic acid you need:

0.03582mol of oxalic acid × (2mol KMnO4 / 5mol Oxalic Acid) = 0.01433mol of KMnO4

If concentration of KMnO4 is 0.250M, liters in 0.01433mol are:

0.01433mol × (1L /0.250mol) = 0.0573L ≡ 57.3mL of 0.250M KMnO4 are required

I hope it helps!

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%