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Chapter 4: Q4.124 CP (page 183)

Sodium peroxide (Na2O2) is often used in self-contained breathing devices, such as those used in fire emergencies, because it reacts with exhaled CO2 to form Na2CO3 and O2. How many liters of respired air can react with 80.0 g of Na2O2 if each liter of respired air contains 0.0720 g of CO2?

Short Answer

Expert verified

The volume of the air is 627.08 L.

Step by step solution

01

Chemical equation 

The balanced chemical equation for the given reaction is,

2Na2O2(s)+2CO2(g)→2Na2CO3(s)+O2(g)

02

Determination of moles

Moles of Na2O2 are,

Moles=massmolar mass=80.0g77.98g/mol=1.026mol

From the above reaction, it is concluded that 2 mol of Na2O2 reacts with 2 mol of CO2.

So, moles of CO2 = 1.026 mol

03

Determination of volume of air

Mass of CO2 is,

Mass=moles×molar mass=1.026mol×44.01g/mol=45.15g

Now, the volume of the air is,

45.15g×1L air0.0720g CO2=627.08L

Thus, the volume of air is 627.08 L.

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