# In the reaction of copper with

Inthe reaction of copper with potassium permanganate and sulfuricacid, the products are copper (II) sulfate, potassium sulfate,water, and manganese (II) sulfate. First, write a balanced netionic equation for this reaction. Then consider the reaction takingplace with 100 grams of copper, 100 grams of potassiumpermanganate, and 400 grams of sulfuric acid. How many grams ofmanganese (II) sulfate could be produced in this reaction?

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

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