Question

In the reaction,
$4N{H}_3+5O_2\to 4NO+6H_{2}O$
when two moles of ammonia and three moles of oxygen are made to react completely, then:

• A. 1 mole of $H_{2}O$ is produced
• B. $N{H}_3$ will be in excess
• C. 2 moles of NO is formed
• D. Oxygen is completely consumed

Answer 1

The correct option is C 2 moles of NO is formed

Finding limiting reagent:
Moles of ammonia = 2 mol
$\frac{\text{moles of ammonia}}{\text{stoichimetric coefficient}}=\frac{2}{4}=0.50$

Moles of $O_2$ = 3
$\frac{O_2}{\text{stoichimetric coefficient}}=\frac{3}{5}=0.6$

Hence, ammonia is the limiting reagent.
all of the ammonia will be consumed(i.e. 2 moles)

Theoretical yield of the reaction,
$\frac{molesofammonia}{4}=\frac{molesofNO}{4}$
Hence, moles of $NO$ formed will be 2.