Question

ratio of active masses of 22g CO2,3g H2,and 7g N2 in a gaseous mixture​

Answer 1

Answer:

Explanation:

Let the volume of gaseous mixture is V

CO  

2

​  

=  

44

22

​  

=0.5moles

H  

2

​  

=  

2

3

​  

=1.5moles

N  

2

​  

=  

28

7

​  

=0.25moles

Total moles = 0.5+1.5+0.25=2.25

Ratio of active masses =0.5/2.25:1.5/2.25:0.25/2.25=1:3:0.5