Question

How many moles of H2 and N2 can be formed by the decomposition of 0.297 mol of ammonia, NH3?

Answer 1

0.4455 mol H₂ and 0.1485 mol N₂

Step-by-step explanation:

Given that,

Moles of ammonia = 0.297

We know that,

The balance equation is,

`2NH_(3)\Rightarrow N_(2)+3H_(2)`

We need to calculate the mole of
`H_(2)`

Using given data

For H₂,

`mole\ of\ H_(2)=0.297\ mol\ of\ NH_(3)*(3\ mol\ H_(2))/(2\ mol\ NH_(3))`

`mole\ of\ H_(2)=0.4455`

For N₂,

`mole\ of\ N_(2)=0.297\ mol\ of\ NH_(3)*(1\ mol\ N_(2))/(2\ mol\ NH_(3))`

`mole\ of\ H_(2)=0.1485`

Hence, 0.4455 mol H₂ and 0.1485 mol N₂