Question

Consider the following reaction:

$\mathbf{4}{\mathbf{NH}}_{\mathbf{3}}\left(\mathbf{g}\right)\mathbf{+}\mathbf{5}{\mathbf{O}}_{\mathbf{2}}\left(\mathbf{g}\right)\to \mathbf{4}\mathbf{NO}\left(\mathbf{g}\right)\mathbf{+}\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{g}\right)$

If a container were to have 10 molecules of ${\mathit{O}}_{\mathbf{2}}$and 10 molecules of ${\mathbf{NH}}_{\mathbf{3}}$initially, how many total molecules (reactants plus products) would be present in the container after this reaction goes to completion?

Short answer

The total molecules present in the container after this reaction goes to completion are 2 molecules of ${\text{NH}}_3$,8 molecules of $NO$and 12 molecules of ${\text{H}}_2\text{O}$.

Step-by-step solution

Definition

The mass of any material expressed in atomic mass units is quantitatively equivalent to its molar mass in grams per mole.

Ratio between NH3 and  O2

$4:5=x:10$

X=Molecules of ${\text{NH}}_3$required

$=\text{8}$

Molecules of ${\text{NH}}_3$in excess$=10-\text{8}$

$=2$

Ratio between O2 and NO

$5:4=10:x$

X=Molecules of $\text{NO}$required

$=\text{8 =}$Molecules of$\text{NO}$

Ratio between O2  and  src="data:image/svg+xml;base64,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" role="math" localid="1648623953829" H2O

$5:6=10:x$

$\text{X}=12=$Molecules of${\text{H}}_2\text{O}$

Therefore2molecules of${\text{NH}}_3$,8molecules of$NO$and12 molecules of${\text{H}}_2\text{O}$.