Respuesta :
Answer:
23.0 %
Explanation:
Let the gaseous mixture has total 100 moles
Composition of [tex]CO_2[/tex] = 0.500 %
The moles of [tex]CO_2[/tex] = [tex]\frac {0.500}{100}\times 100\ moles[/tex] = 0.5 moles
Also,
Molar mass of carbon dioxide = 44.01 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]0.5\ moles= \frac{Mass}{44.01\ g/mol}[/tex]
Mass of [tex]CO_2[/tex] = 22.005 g
Composition of air = 99.500 %
The moles of air = [tex]\frac {99.500}{100}\times 100\ moles[/tex] = 99.500 moles
Also,
Given, Average molar mass of air = 29.0 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]99.5\ moles= \frac{Mass}{29.0\ g/mol}[/tex]
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 [tex]O_2[/tex] = 21 % of air
The moles of [tex]O_2[/tex] = [tex]\frac {21}{100}\times 99.5\ moles[/tex] = 20.895 moles
Also,
Molar mass of oxygen = 31.9988 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]20.895\ moles= \frac{Mass}{ 31.9988\ g/mol}[/tex]
Mass of [tex]O_2[/tex] = 668.615 g
Thus,
% O by mass = [tex]\frac {Mass\ of\ Oxygen}{Total\ mass}\times 100=\frac {668.615}{2907.505}\times 100[/tex] = 23.0 %
