The Occupational Safety and Health Administration has set the limit on the maxim

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.

Explanation / Answer

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 %

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.