Question

Potassium permanganate, KMnO4, a common oxidizing agent, is made from various ores that contain manganese(IV) oxide, MnO2. The following equation shows the net reaction for one process that forms potassium permanganate.

a. 2MnO2 +2KOH+O2 → 2KMnO4 +H2 What is the maximum mass, in kilograms, of KMnO4 that can be made from the reaction of 453 g of MnO2 with 943 g of KOH and excess oxygen gas?

b.Explain why the oxygen gas is in excess.

c. If 1.18 kg of KMnO4 are isolated in the reaction above, what is the percent yield of KMnO4?

Answer 1

(a) 2MnO2 +2KOH+O2 → 2KMnO4 +H2

mass of MnO2 = 453 g

mass of KOH = 943 g

oxygen is in excess amount.

moles of MnO2 = mass / molar mass = 453g / 86.94 g/mol = 5.21 mol

moles of KOH = mass / molar mass = 943 g / 56.10 g/mol = 16.81 mol

From the balanced equation,

Therefore MnO2 is limiting reagent.

Moles of KMnO4 can be obtained = 5.21 mol

Amount of KMnO4 can be obtained = 5.21 mol x 158.03 g/mol = 823 g = 0.823 kg

(b) Oxygen gas is in excess becauseall the MnO2 has to be oxidised to KMnO4.

(c) Percent yield of KMnO4 = (actual yield / theoretical yield ) x 100

Percent yield of KMnO4 = ( 1.18 kg / 0.823 kg ) x 100 = 143 %