Answer to Question #280586 in General Chemistry for jhon

Question

Q7. A gas mixture that has the following composition in percent by volume: 10.2% CO2,

21.0% CO, 18.5% H2, 2.3% CH4 and 48.0% N2.

(a) What is the density of this gas at 24 °C and 755 mmHg, in grams per liter?

(b) What is the partial pressure of CO in this mixture at STP?

(c) What volume of air is required for the complete combustion of this mixture of gas? (Air

contains 21% O2 by volume)

Answer 1

Density of the mixture

Molecular mass of the mixture

(10.2% X44) + (18.5% X 2) + (21% X28) + (2.3% X 16) + (48% X14)

4.488 +0.37 + 5.88 + 0.368 +6.72

=17.862gmol-1

Now apply Charles law and Boyles law of gases at STP

273/297 X 755/760 X 17.826

=16.277gmol-1 is the density of the mixture

Partial pressure of CO in the mixture at STP

At STP pressure of the mixture is 17.862gmol-1

18.5% X 17.862

=3.298gmol-1 is the partial pressure of CO at STP

Volume of air required to completely combust the mixture

One mole of all gases contains the same number of particles according to avogadros

But they contain different molar masses

For oxygen gas it is 32 and for mixture it is 17.862

17.862/32

=0.5571 L of oxygen can be used to combust the mixture.

This is only 21% of air so volume of air will be

0.5571 X 100/21

=2.652L of Air