Q4.124 CP - Chemistry: Molecular Nature Of Matter And Change

Question

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?

Step-by-Step Solution

Answer

The volume of the air is 627.08 L.

Step 1: Chemical equation

The balanced chemical equation for the given reaction is,

 

 $\mathbf{2}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

Step 2: Determination of moles

Moles of Na₂O₂ are,

 

 $\begin{array}{c}\mathbf{Moles}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{molar}\mathbf{mass}}\\ \mathbf{=}\frac{\mathbf{80}\mathbf{.0}\mathbf{g}}{\mathbf{77}\mathbf{.98}\mathbf{g}\mathbf{/}\mathbf{mol}}\\ \mathbf{=}\mathbf{1}\mathbf{.026}\mathbf{mol}\end{array}$

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

 

Step 3: Determination of volume of air

Mass of CO₂ is,

 $\begin{array}{c}\mathbf{M}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{=}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{s}\mathbf{\times }\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\\ \mathbf{=}\mathbf{1}\mathbf{.026}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{\times }\mathbf{44}\mathbf{.01}\mathbf{g}\mathbf{/}\mathbf{m}\mathbf{o}\mathbf{l}\\ \mathbf{=}\mathbf{45}\mathbf{.15}\mathbf{g}\end{array}$

 

Now, the volume of the air is,

 $\mathbf{45}\mathbf{.15}\mathbf{g}\mathbf{\times }\frac{\mathbf{1}\mathbf{L}\mathbf{air}}{\mathbf{0}\mathbf{.0720}\mathbf{g}{\mathbf{CO}}_{\mathbf{2}}}\mathbf{=}\mathbf{627}\mathbf{.08}\mathbf{L}$

 

Thus, the volume of air is 627.08 L.