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.

A) What mass of carbon dioxide is present in 1.00 m3 of dry air at a temperature of 21 C and a pressure of 703 torr ?

B) Calculate the mass percentage of oxygen in dry air.

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

we know that

for gases

PV = nRT

now

pressure = 703 torr = 703/760 atm

volume = 1 m3 = 1000 L

temperature (T) = 21 C = 21 + 273 = 294 K

R = 0.0821

so

(703 / 760) x 1000 = n x 0.0821 x 294

n = 38.3227

so

moles of air present = 38.3227

now

mole fraction of C02 = moles of C02 / moles of air

given

mole fraction of C02 = 0.000375

so

0.000375 = moles of C02 / 38.3227

moles of C02 = 0.014370851

now

mass = moles x molar mass

molar mass of C02 = 44.0099 g/mol

so

mass of C02 present = 0.014370851 x 44.0099

mass of C02 present = 0.63246 g

so

0.63246 grams of C02 is present in 1 m3 of dry air

B)

now

mass of air = mass of N2 + mass of 02 + mass of Ar + mass of C02

now

moles of N2 = 0.78084 x 38.3227 = 29.923897

mass of N2 = 29.923897 x 28.013

mass of N2 = 838.258

now

moles of 02 = 0.20948 x 38.3227 = 8.027839

mass of 02 = 8.027839 x 32

mass of 02 = 256.891

now

moles of Ar = 0.00934 x 38.3227 = 0.35793

mass of Ar = 0.35793 x 39.948

mass of Ar = 14.3 g

so

mass of air = 838.258 + 256.891 + 14.3 + 0.63246

mass of air = 1110.08146

now

mass percent of oxygen = mass of oxygen x 100 / mass of air

mass percent of oxygen = 256.891 x 100 / 1110.08146

mass percent of oxygen = 23.14

so

mass percent of oxygen in dry air is 23.14 %

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