# Given that 4 NH3+5 O2 --> 4 NO + 6H2O, when 4.50 mol of H2O are formed, the amount of NO formed is?

Apr 8, 2018

The given balanced equation is,

$4 N {H}_{3} + 5 {O}_{2} \to 4 N O + 6 {H}_{2} O$

This implies, $4 \text{ moles}$ of $N {H}_{3}$ reacts with $5 \text{ moles}$ of ${O}_{2}$ to produce $4 \text{ moles}$ of $N O$ and $6 \text{ moles}$ of ${H}_{2} O$

So, the net products will be, $4 \text{ moles}$ of $N O$ and $6 \text{ moles}$ of ${H}_{2} O$

Accordingly, $x \text{ moles}$ of $N O$ and $4.5 \text{ moles}$ of ${H}_{2} O$ should be formed.

where $x$ would be,
$x = \frac{4 \times 4.5}{6}$

$x = 3 \text{ moles}$

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Note : The $4 : 6 \implies 2 : 3$ ratio remains constant for the given pair of products.
As from the question, if the answer $3 \text{ moles}$ is correct, then $3 : 4.5$ must be equal to $2 : 3$

Checking that, $3 : \frac{9}{2} \implies 6 : 9 \implies 2 : 3$

therefore, the answer must be correct.

$3 \text{ moles}$ of $N O$ is formed along with $4.50 \text{ moles}$ of ${H}_{2} O$.