Question

Consider the following balanced reaction between hydrogen and nitrogen to form ammonia:3H2(g) + N2(g)→2NH3(g)How many moles of NH3 can be produced from 24.0 mol of H2 and excess N2?Express the number of moles to three significant figures.

Answer 1

16.0 moles

Step-by-step explanation:

The reaction is:

`3H_2_((g))+N_2_((g))\rightarrow 2NH_3_((g))`

It is given that
`N_2` is present in large excess. It means that
`H_2` is the limiting reagent and the formation of the product is governed by
`H_2`.

From the reaction stoichiometry,

3 moles of
`H_2` on reaction forms 2 moles of
`NH_3`

Also,

1 mole of
`H_2` on reaction forms 2/3 moles of
`NH_3`

Given moles = 24.0 moles

24 moles of
`H_2` on reaction forms (2/3)×24.0 moles of
`NH_3`

Moles of
`NH_3` = 16.0 moles