Question

Problem 3.62

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. (Figure 1)

Part A

How many moles of O2 are produced when 0.350 mol of KO2 reacts in this fashion?

Express your answer with the appropriate units.

n(O2) =

0.263 mol

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Correct

Part B

How many grams of KO2 are needed to form 6.0 g of O2?

Express your answer with the appropriate units.

m_(KO2) =

24.9g

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Incorrect; Try Again; 4 attempts remaining

Part C

How many grams of CO2 are used when 6.0 g of O2 are produced?

Express your answer with the appropriate units.

m_(CO2) =

Answer 1

A)
From reaction,
moles of O2 = (3/4)*moles of KO2
= (3/4)*0.350 moles
= 0.263 mol

B)

mass of O2 = 6 g
molar mass of O2 = 32 g/mol
mol of O2 = (mass)/(molar mass)
= 6/32
= 0.1875 mol

  

According to balanced equation
mol of KO2 required = (4/3)* moles of O2
= (4/3)*0.1875
= 0.25 mol



mass of KO2 = number of mol * molar mass
= 0.25*71.1
= 17.8 g
Answer: 17.8 g

C)

mass of O2 = 6 g
molar mass of O2 = 32 g/mol
mol of O2 = (mass)/(molar mass)
= 6/32
= 0.1875 mol


According to balanced equation
mol of CO2 required = (2/3)* moles of O2
= (2/3)*0.1875
= 0.125 mol



mass of CO2 = number of mol * molar mass
= 0.125*44.01
= 5.50 g
Answer: 5.50 g