Please explain how to solve using Gay Lussac's law of combining volumes: 

1)20ml. each of oxygen and hydrogen and 10 ml. of carbon monoxide are exploded in an enclosure .What will be the volume and composition of the mixture of the gases when they are cooled to room temperature. 

2)100cc each of water gas and oxygen are ignited and the resultant mixture of gases cooled to room temperature.Calculate the composition of the resultant mixture.
 

Asked by manalip2505 | 26th Jan, 2019, 11:14: AM

Expert Answer:

1) 

Given:

Volume of O2 VO2 = 20 ml

 

Volume of H2 VH2 = 20 ml

 

Volume of CO VCO = 10 ml

 

Oxygen and hydrogen are exploded in an enclosure;

 

2H2 + O2 → 2H2O  ......(1)

 

Oxygen and carbon monoxide are exploded in an enclosure;

 

2CO + O2 → 2CO2 ........(2)

 

From Eq (1) 

 

2 V of H2 reacts with 1 V of the O2.

 

Therefore, 20 ml of H2 reacts with 10 ml of O2.

 

So, 20- 20 = 10 ml of O2 remains unreacted.

 

From Eq (2) 

 

2 V of CO reacts with 1 vol of O2 

 

Therefore, 10 ml of CO reacts with 5 ml of  O2 and gives 10 ml of CO2 

 

So, from 10 ml of O2 10-5 = 5 ml remains unreacted.

 

The total composition of the resulting mixture is 5 ml O2 and 10 ml CO2.

 

2)

CO subscript left parenthesis straight g right parenthesis end subscript space plus space straight H subscript 2 subscript left parenthesis straight g right parenthesis end subscript space plus space straight O subscript 2 subscript left parenthesis straight g right parenthesis end subscript space rightwards arrow CO subscript 2 subscript left parenthesis straight g right parenthesis end subscript space plus space straight H subscript 2 straight O
1 vol space space space space space 1 vol space space space space space space 1 vol space space space space space space space 1 vol space space space space space space space space 1 vol

Volume space of space water space gas equals 100 cc

Volume space of space CO equals 50 cc

Volume space of space straight H subscript 2 equals 50 cc space

Volume space of space straight O subscript 2 equals 100 cc space

2 CO subscript left parenthesis straight g right parenthesis end subscript space plus space straight O subscript 2 subscript left parenthesis straight g right parenthesis end subscript space rightwards arrow 2 CO subscript 2 subscript left parenthesis straight g right parenthesis end subscript space

By space Gay minus Lussac apostrophe slaw comma

1 vol space o straight f space space CO space requires space 1 space vol space of space oxygen

therefore 50 space cc space of space CO space will space produce space 50 space cc space of space CO subscript 2 space

After space reaction comma space the space composition space of space gaseous space mixture space will space be comma

2 straight H subscript 2 subscript left parenthesis straight g right parenthesis end subscript space plus space straight O subscript 2 subscript left parenthesis straight g right parenthesis end subscript space rightwards arrow 2 straight H subscript 2 straight O

Amount space of space straight O subscript 2 space used equals space 50 cc space

CO subscript 2 equals 50 cc space

Oxygen equals 100 minus 50
space space space space space space space space space space space space equals 50 cc

 

Answered by Varsha | 29th Jan, 2019, 12:35: PM