Question

The Occupational Safety and Health Administration has set the limit on the maximum percentage of carbon dioxide (CO2) in air a worker can breathe at 0.500% by mole. If dry ice (solid CO2) is sublimating in a room at a rate of 0.610 mol/min, what is the minimum rate that fresh air (mol air/min) has to be supplied to the room so that the concentration of CO2 in the room doesn’t exceed 0.500% by mole? Assume the room is well mixed, and the volume of the solid dry ice does not change.

What is the percent by mass of oxygen in a gaseous mixture whose molar composition is 0.500% CO2 and 99.500% air? The composition of air is 21% mole O2, 70% mole N2 and has an average molar mass of 29.0 g/mol.

Answer 1

Solution :-

Given data

Rate flow in of the CO2

Rate of CO2 = 0.610 mol / min

Concentration of the CO2 needed = 0.500 % by mole

So lets calculate the moles of the air needed to get this concentration of the CO2

Total moles of gases in the room are calculated as

0.610 mol CO2 * 100 % /0.500 % = 122 mol

So total moles of gases = 122 mol

Now lets calculate the moles of the air

Moles of air = total moles   -   moles of CO2

                       = 122 mol – 0.610 mol

                      = 121.4 mol air

So the minimum rate of the air flow needs to 121.4 mol air / min

Part 2

% of the N2 should be 79 % so that we get the 100 % of the total percentage.

0.500 % CO2 * 1 mol / 100 % = 0.005 mol CO2

Molar mass of 99.5 % air = 29 g per mol * 95.5 % / 100 % = 27.695 g per mol

Now lets calculate the mass of CO2

Mass of CO2 = 0.005 mol * 44.01 g per mol = 0.22005 g

Total mass of Air + CO2 = 27.695 g + 0.22005 g = 27.91505 g per mol

Now lets calculate the moles of the O2

Moles of O2 = 21 % * 1 mol / 100 % = 0.21 mol O2

Mass of O2 = 0.21 mol * 32.0 g per mol = 6.72 g

Now lets calculate the percent by mass of O2

% mass of O2 = (mass of O2 / molar mass of air )*100%

                       = (6.72 g / 27.91505 g)*100%

                         = 24.07 %

So the mass percent of the O2 is 24.07 %