Question

Sodium peroxide is used in orbital stations for air purification.The reaction that takes place is?

Sodium peroxide is used in orbital stations for air purification.The reaction that takes place is:
`Na_2O_2+CO_2=Na_2CO_3+\frac{1}{2}O_2`

An orbital station has the volume of 220m^3.At a certain point CO2 concentration has reached 20%.What is the mass of 75% purity sodium peroxide that should have been used to lower CO2 concentration to 1%(in volumes),in normal conditions (T=273K , p=1atm) , respectively what is the volume of oxygen regenerated?

Answer 1

Warning! Long Answer. You must use 190 kg of the impure sodium peroxide. You regenerate `"21 m"^3` of `"O"_2`.

Explanation:

Step 1. Calculate the volume of `"CO"_2` that must be removed

You want to reduce the volume of `"CO"_2` from 20 % to 1 % of the volume of the space station (i.e. by 19 %).

`"Volume of CO"_2 = 220 color(red)(cancel(color(black)("m"^3))) × ("19 m"^3color(white)(l) "CO"_2)/(100color(red)(cancel(color(black)("m"^3)))) = "41.8 m"^3color(white)(l) "CO"_2`

Step 2. Calculate the moles of `"CO"_2` to be removed

We can use the Ideal Gas Law for this calculation.

`color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "`

We can rearrange this formula to get

`n = (pV)/(RT)`

∴ `n = (1 color(red)(cancel(color(black)("atm"))) × "41 800" color(red)(cancel(color(black)("L"))))/("0.082 06"color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 273color(red)(cancel(color(black)("K")))) = "1866 mol CO"_2`

Step 4. Calculate the moles of pure `"Na"_2"O"_2` that must react with the `"CO"_2`

The balanced equation is

`"Na"_2"O"_2 + "CO"_2 → "Na"_2"CO"_3 + ½"O"_2`

`"Moles of Na"_2"O"_2 = 1866 color(red)(cancel(color(black)("mol CO"_2))) × (1 "mol Na"_2"O"_2)/(1 color(red)(cancel(color(black)("mol CO"_2)))) = "1866 mol Na"_2"O"_2`

Step 5. Calculate the mass of pure `"Na"_2"O"_2` that must react with the `"CO"_2`

`"Mass of Na"_2"O"_2 = 1866 color(red)(cancel(color(black)("mol Na"_2"O"_2))) × ("77.98 g Na"_2"O"_2)/(1 color(red)(cancel(color(black)("mol Na"_2"O"_2)))) = "145 500 g Na"_2"O"_2 = "145.5 kg Na"_2"O"_2`

Step 6. Calculate the mass of impure `"Na"_2"O"_2`

The `"Na"_2"O"_2` is only 75 % pure, so you will have to use more than 146 kg of it.

`"Mass of impure Na"_2"O"_2 = 145.5 color(red)(cancel(color(black)("kg Na"_2"O"_2))) × ("100 kg impure Na"_2"O"_2)/(75 color(red)(cancel(color(black)("kg Na"_2"O"_2)))) = "190 kg Na"_2"O"_2` (2 significant figures)

Note: The answer can have only two significant figures, because that is all you gave for the percent of carbon dioxide and for the purity of the sodium peroxide.

Step 7. Calculate the volume of `"O"_2` regenerated

`"Na"_2"O"_2 + "CO"_2 → "Na"_2"CO"_3 + ½"O"_2`

Thus, the removal of 1 mol of `"CO"_2` results in the formation of 0.5 mol of `"O"_2`.

We can also say, "The removal of `"1 m"^3` of `"CO"_2` results in the formation
of `"0.5 m"^3` of `"O"_2`".

Thus, the removal of `"41.8 m"^3` of `"CO"_2` results in the regeneration
of `"21 m"^3` of `"O"_2` (2 significant figures).