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Q.
$\underset{20\, g}{N _{2( g )}}+\underset{5 \,g}{3 H _{2( g )}} \rightleftharpoons 2 NH _{3( g )}$
Consider the above reaction, the limiting reagent of the reaction and number of moles of $NH _3$ formed respectively are:
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Solution:
$ \underset{W _2=20\, g}{N _2( g )}+\underset{5\, g}{3 H _2( g )} \rightleftharpoons 2 NH _3( g )$
$ n =\frac{20}{28} \frac{5}{2}$
Stoichiometric Amount:
$N _2 \rightarrow \frac{20 / 28}{1}=\frac{20}{28} H _2 \rightarrow \frac{5 / 2}{3}=\frac{5}{6}$
$\therefore N _2$ is the Limiting Reagent.
$\therefore n \left( NH _3\right) =2 \times n \left( N _2\right)=2 \times \frac{20}{28} $
$=1.42$