Question

12.33 L of H2 and 36.99 L of N2 are mixed at certain temperature and pressure to prepare NH3. The volume of NH3 produced at same temperature and pressure is ​

Answer 1

Answer:

7.46L

Explanation:

N2 + 3H2-- 2NH3

moles of H2= 12.33/22.4

= 0.5

moles of N2= 36.99/22.4

=1.6

therefore limiting reagent here is H2

3 moles of H2 --- 2 moles of NH3

0.5 moles of H2-- 1/3 moles of NH3

no.of moles=1/3

1/3= vol. required/22.4

vol. required=22.4/3

=7.46 L

hope u find ur answer in it