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Dry air is essentially a mixture of the following entities: N2,O2,Ar\mathrm{N}_2, \mathrm{O}_2, \mathrm{Ar}, and CO2\mathrm{CO}_2. The composition of dry air, in mole percent, is 78.08% N2,20.95%O2,0.93%Ar78.08 \% \mathrm{~N}_2, 20.95 \% \mathrm{O}_2, 0.93 \% \mathrm{Ar}, and 0.04%CO20.04 \% \mathrm{CO}_2. What is the mass, in grams, of a sample of air that contains exactly one mole of the entities?

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Answered 2 years ago
Answered 2 years ago

Find the mass in grams of a sample of air.

(1) First, list the molar masses of N2\mathrm{N_{2}}, O2\mathrm{O_{2}}, Ar, and CO2\mathrm{CO_{2}} as shown below:

  • Molar mass of N2\mathrm{N_{2}} = 28.02 g
  • Molar mass of O2\mathrm{O_{2}} = 32.00 g
  • Molar mass of Ar\mathrm{Ar} = 39.95 g
  • Molar mass of CO2\mathrm{CO_{2}} = 44.01 g

(2) Then, calculate for the mass of air that contains of each mole of entities as shown below:

Mass of air = (78.08100)(\dfrac{78.08}{100}) mol N2\mathrm{N_{2}} x 28.02 g1 molN2\dfrac{\text{28.02 g}}{\text{1 mol}\mathrm{N_{2}}} + (20.95100)(\dfrac{20.95}{100}) mol O2\mathrm{O_{2}} x 32.00 g1 molO2\dfrac{\text{32.00 g}}{\text{1 mol}\mathrm{O_{2}}} + (0.93100)(\dfrac{0.93}{100}) mol Ar\mathrm{Ar} x 39.95 g1 molAr\dfrac{\text{39.95 g}}{\text{1 mol}\mathrm{Ar}} + (0.04100)(\dfrac{0.04}{100}) mol CO2\mathrm{CO_{2}} x 44.01 g1 molCO2\dfrac{\text{44.01 g}}{\text{1 mol}\mathrm{CO_{2}}}

Mass of air = 28.97 g

Thus, the mass of air in this given problem is 28.97.

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