In the reaction, 4N H_3 _ (g) + 5 O_2 _ (g) 4NO_ (g) + 6 H_2 O_ (…

MCQ

In the reaction, $4N{H_3}_{(g)} + 5{O_2}_{(g)} \to 4NO_{(g)} + 6{H_2}O_{(g)}$, When $1$ mole of ammonia and $1$ mole of ${O_2}$ are made to react to completion

A
$1.0$ mole of ${H_2}O$ is produced
B
$1.0$ mole of $NO$ will be produced
✓
All the oxygen will be consumed
D
All the ammonia will be consumed

✓

Answer

Correct option: C.

All the oxygen will be consumed

c
(c)$\begin{array}{*{20}{c}}{4N{H_{3(g)}} + 5{O_{2(g)}} \to 4N{O_{(g)}} + 6{H_2}{O_{(g)}}\,}\\{t = 0\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\,t = t\,\,\,\,\,\,1 - 4x\,\,\,\,1 - 5x\,\,\,\,\,\,\,\,4x\,\,\,\,\,\,\,\,\,\,\,\,\,\,6x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}$

Oxygen is limiting reagent

So, $X = \frac{1}{5} = 0.2$ all oxygen consumed

Left $N{H_3} = 1 - 4 \times 0.2 = 0.2$.

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