In the reaction of copper with

Cu + KMnO4 + H2SO4 CuSO4 + 2MnSO4 + K2SO4 + 8 H2O

The balanced equation is5 Cu + 2 KMnO4 + 8 H2SO4 5 CuSO4 + 2MnSO4 + K2SO4 + 8 H2O

In the above balanced reaction above

5 mole of Cu reacts with 2 moles of KMnO4 and 8 moles of H2SO4to produce 5 mole of CuSO4, 2 moles of MnSO4, 1 mole of K2SO4 and 8mole H2O.

100 grams of copperMolar mass of Cu = 63.5 g/molSo, 63.5 g of Cu = 1 mol 1 g of Cu =(1/63.5) mol 100 g of Cu= (100/63.5) mol = 1.57 mol

100 grams of potassium permanganateMolar mass of KMnO4 = 158 g/molSo, 158 g of KMnO4 = 1 mol 1 g ofKMnO4 = (1/158) mol 100 g ofKMnO4 = (100/158) mol = 0.63 mol

400 grams of sulfuric acidMolar mass of H2SO4 = 98 g/molSo, 98 g of H2SO4 = 1 mol 1 g ofH2SO4 = (1/98) mol 100 g ofH2SO4 = (100/98) mol = 1.02 mol

Now,Cu = 1.57 mol / 5 = 0.314 mol (Limiting reagent)KMnO4 = 0.63 mol / 2 =  0.315 molH2SO4 = 1.02 mol / 8 = 0.128 molAnyway, H2SO4 is used as a catalytic amount in the reaction.

Again,5 Cu + 2 KMnO4 + 8 H2SO4 5 CuSO4 + 2MnSO4 + K2SO4 + 8 H2O

In the above balanced reaction above

5 mole of Cu reacts to produce 2 moles of MnSO4. 1 mole ofCu reacts to produce 2/5 moles of MnSO4. 1.57 moleof Cu reacts to produce 1.57 x (2/5) moles of MnSO4. 1.57 moleof Cu reacts to produce 0.628 moles of MnSO4.

Molar mass of MnSO4 = 151 g/molSo, 1 mole of MnSO4 = 151 g 0.628 moleof MnSO4 = 0.628 x 151 g = 94.83 g