Air is a mixture of several gases. The 10 most abundant of these gases are liste

Question

Air is a mixture of several gases. The 10 most abundant of these gases are listed here along with their mole fractions and molar masses.

Part A

What mass of carbon dioxide is present in 1.00 m3 of dry air at a temperature of 23?C and a pressure of 695torr ?

Part B

Calculate the mass percentage of oxygen in dry air.

Express your answer with the appropriate units.

44.0128

Component Mole fraction Molar mass
(g/mol) Nitrogen 0.78084 28.013 Oxygen 0.20948 31.998 Argon 0.00934 39.948 Carbon dioxide 0.000375 44.0099 Neon 0.00001818 20.183 Helium 0.00000524 4.003 Methane 0.000002 16.043 Krypton 0.00000114 83.80 Hydrogen 0.0000005 2.0159 Nitrous oxide 0.0000005

44.0128

Explanation / Answer

a) 1.00 m^3=1 L
739 torr = 739/760 atm
25 C= 298 K

using PV=nRT , n =PV/RT = 39.2 mole
so CO2 in mass = 39.2 * Cf * 44.0099=64.7g

pv =nRT

pv= wRT/M

W = pvm/RT

   739*1*44/760*0.0821*298 =1.74g

b) total mass of 1 mole air=(28.013*0.78084++44.0128*0000005)=28... g
in 1 mole O2 is=.20948 mole=6.7029g

so percentage of O2 as mass = 6.7027*100%/ 28.97 =23.14%

Assuming 100 molecules, 21 will be oxygen, 78 will be nitrogen and almost 1 will be argon. We can bump that up to moles so as to get grams, by multiplying by Avogadro's constant.

21.0 mol O2 x (32.0g O2 / 1 mol O2) = 672 grams O2
78.0 mol N2 x (28.0g N2 / 1 mol N2) = 2184g N2
1 mol Ar x (39.9g Ar / 1 mol Ar) = 39.9g Ar

Total mass = 2896g
mass % O2 = 672g / 2896g x 100 = 23.2% O2
mass % N2 = 2184g / 2896g x 100 = 75.4% N2
mass % Ar = 40.g / 2896g x 100 = 1.4% Ar

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