Question

In the reaction of copper with potassium permanganate and sulfuric acid, the products are copper (II) sulfate, potassium sulfate, water, and manganese (II) sulfate. First, write a balanced net ionic equation for this reaction. Then consider the reaction taking place with 100 grams of copper, 100 grams of potassium permanganate, and 400 grams of sulfuric acid. How many grams of manganese (II) sulfate could be produced in this reaction?

Answer 1

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

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

In the above balanced reaction above

5 mole of Cu reacts with 2 moles of KMnO4 and 8 moles of H2SO4 to produce 5 mole of CuSO4, 2 moles of MnSO4, 1 mole of K2SO4 and 8 mole H2O.

100 grams of copper
Molar mass of Cu = 63.5 g/mol
So, 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 permanganate
Molar mass of KMnO4 = 158 g/mol
So, 158 g of KMnO4 = 1 mol
1 g of KMnO4 = (1/158) mol
100 g of KMnO4 = (100/158) mol = 0.63 mol

400 grams of sulfuric acid
Molar mass of H2SO4 = 98 g/mol
So, 98 g of H2SO4 = 1 mol
1 g of H2SO4 = (1/98) mol
100 g of H2SO4 = (100/98) mol = 1.02 mol

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

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

In the above balanced reaction above

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

Molar mass of MnSO4 = 151 g/mol
So, 1 mole of MnSO4 = 151 g
0.628 mole of MnSO4 = 0.628 x 151 g = 94.83 g