Question

The reaction between potassium superoxide, KO2, and CO2, 4KO2+2CO2→2K2CO3+3O2
is used as a source of O2 and absorber of CO2 in self-contained breathing equipment used by rescue workers.

a). How many moles of O2 are produced when 0.475 mol of KO2 reacts in this fashion?

b). How many grams of KO2 are needed to form 8.0 g of O2?

c). How many grams of CO2 are used when 8.0 g of O2 are produced?

Answer 1

4KO2 + 2CO2 ------> 2K2CO3 + 3O2

a) Stoichiometrically ,3moles of O2 is produced from 4moles of KO2

Therefore

Number of moles of O2 produced by 0.475moles of KO2 = (3/4) ×0.475mol = 0.3563mol

b)

number of moles of O2 needed = 8.0g/32g/mol = 0.0.25mol

moles of KO2 needed = (4/3) × 0.25mol = 0.3333mol

mass of KO2 required = 0.3333mol × 71.1g/mol = 23.70g

c)

number of moles of O2 produced = 0.25mol

number of moles of CO2 required = (2/3)× 0.25mol = 0.1667mol

mass of CO2 required = 0.1667mol × 44.01g/mol = 7.336g