Question

40% of a mixture of 0.2 mol of n2 and 0.6 mol of h2 reacts to give nh3 according to the equation : n (g)+3h (g) 2nh (g) â 2 2 3 at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are

Answer 1

Hey mate,

# Answer- 4:5

# Explaination-

For given reaction

N2 + 3H2 ---> 2NH3

(0.2-x) (0.6-3x) 2x

Also 40% reactants burnt, hence 60% left

Now

(0.2-x)+(0.6-3x)=60%×(0.2+0.6)

Solving this we'll get,

x=0.08

Now initial moles

n1=0.2+0.6=0.8

And final moles n2=(0.2-0.08)+(0.6-3*0.08)+(2*0.08) = 0.64

Lets calculate,

final volume/initial volume

= final moles/initial moles

= 0.64/08

= 0.8

Ratio of final volume to initial volume is 0.8=4/5.

Thanks for asking...

Now The ratio of mol=volume=0.64/0.8=4/5(You have to notice 40 percent reacted and 60 percent left )

Answer 2

Answer:

4:5

Explanation:see the photo