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, 79 % mole n2 , and has an average molar mass of 29.0 g/mol number % o by mass

23.0 %

Explanation:

Let the gaseous mixture has total 100 moles

Composition of = 0.500 %

The moles of = = 0.5 moles

Also,

Molar mass of carbon dioxide = 44.01 g/mol

The formula for the calculation of moles is shown below:

Thus,

Mass of = 22.005 g

Composition of air = 99.500 %

The moles of air = = 99.500 moles

Also,

Given, Average molar mass of air = 29.0 g/mol

The formula for the calculation of moles is shown below:

Thus,

Mass of air = 2885.5 g

Total mass  of 100 moles of the mixture = 22.005 g + 2885.5 g = 2907.505 g

Composition of = 21 % of air

The moles of = = 20.895 moles

Also,

Molar mass of oxygen =  31.9988 g/mol

The formula for the calculation of moles is shown below:

Thus,

Mass of = 668.615 g

Thus,

% O by mass = = 23.0 %

elements

explanation:

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