Question

The reaction of potassium superoxide, KO2, is used in life-support systems to replace CO2(g) in expired air with O2(g).
4KO2(s)+2CO2(g)→2K2CO3(s)+3O2(g)

a) How many moles of O2(g) are produced by the reaction of 100 g CO2(g) with excess KO2(s)?

b) How many grams of KO2(s) are consumed per 1000.0 g CO2(g) removed from expired air?

c) How many moles K2CO3 are produced per milligram of KO2 consumed?

Answer 1

a)  Notice that in the equation for every 2 mol of CO2 used up you produce 3 mol of O2. Or 1.5 times more

So in 100 g of CO2 you have to use the formula n = m /M ,where M is molar mass of CO2

100 g / 44.01 g/'mol = 2..27 mol

Therefore the number of mol of O2 produced will be 2.27 x 1.5 = 3.405 mol

b) 1000 gCO2 * 4KO2(71.10 g/mol)/2CO2(44.01g/m)=3231 gKO2

c) 4 mol of KO2 produce 2 mol of K2CO3

Hence 1 mol of KO2 wil produce 0.5 mol of K2CO3

mols of KO2 consumed = 10^-3 g / 71.1 g/mol = 1.4*10^-5

moles of K2CO3 produced = 0.5 * 1.4*10^-5 = 7*10^-6 moles