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.

What mass of carbon dioxide is present in 1.00 m3 of dry air at a temperature of 25 C and a pressure of 737 torr ?

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

a)

mass % of Oxygen in dry air

x-O2 = mass of O2 / total mass

assume a basis of 100 mol of air..

then...

mol of O2 = y-O2 * total mol of air = 0.20948 * 100 = 20.948 mol of O2

mass of O2 = mol*MW = 20.948 * 32 = 670.336 g

MW of air = 28.966 g/mol approx

mass of air = 100*28.966 = 2896.6 gof air

now...

x-O2 = 670.336 / (2896.6 ) = 0.23142 for O2 per mass

b)

mass of CO2 in V = 1 m3... at T = 25°C, 298K at P = 737 torr

so...

let us calculate mol of CO2:

total mol:

PV = nRT

P = 737 / 760 = 0.96973 atm

V =1 m3 = 1000 L

R = 0.082 Latm/molK

T = 298 K

n = PV/(RT) = 0.96973 *(1000)/(0.082*298)

n = 39.6844 mol of air

of which, we know that

y-CO2 = 0.000375

so

mol of CO2 = y-CO2 * total mol = 0.000375*39.6844 = 0.01488 moles of CO2

mass of CO2 = mol*W = 0.01488*44.0099 = 0.6548673 g = 0.6548673*10^3 mg = 654.8673 mg

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