Question

The moles of $O _{2}$ required for reacting with $6.8\, g$ ammonia. $\left(\ldots . . NH _{3}+\ldots . O _{2} \rightarrow \ldots . NO +\ldots . H _{2} O \right)$ is:

• A. 5
• B. 2.5
• C. 1
• D. 0.5

Answer 1

Answer: (D)

The balanced equation is:

$4 NH _{3}+5 O _{2} \rightarrow 4 NO +6 H _{2} O$ ...(1)

$4 \times 17\, g$ of $NH _{3}$ ( eq -1)

Required $O _{2}=5 \,mol$

$6.8$ of $NH _{3}$ require $O _{2}=\frac{5 \times 6.8}{4 \times 17}$

$=0.5 \,mol$