Question

In the reaction,
$4N{H}_3+5O_2\to 4NO+6H_{2}O$,
when one mole of ammonia and 1.5 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. 1 mole of NO is formed
• D. Oxygen is completely consumed

Answer 1

The correct option is C 1 mole of NO is formed

Finding limiting reagent:
Moles of ammonia = 1 mol
$\frac{molesofammonia}{stoichiometriccoefficient}=\frac{1}{4}=0.25$

Moles of $O_2$ = 1.5
$\frac{molesof{O}_2}{stoichiometriccoefficient}=\frac{1.5}{5}=0.3$

Hence, ammonia is the limiting reagent.

Theoretical yield of the reaction,
$\frac{molesofammonia}{4}=\frac{molesofNO}{4}$
Hence, moles of NO formed will be 1.