Elemental phosphorus occurs as tetratomic molecules, ${\mathit{P}}_{\mathbf{4}}\mathbf{.}$What mass of chlorine gas is needed to react completely with 455 g of phosphorus to form phosphorus pentachloride?

When tetratomic molecules, ${\mathit{P}}_{\mathbf{4}}$reacts with $\mathit{C}{\mathit{l}}_{\mathbf{2}}$and forms $\mathit{P}\mathit{C}{\mathit{l}}_{\mathbf{5}}\mathbf{.}$

The chemical reaction is:

${\mathit{P}}_{\mathbf{4}}\left(s\right)\mathbf{+}\mathit{C}{\mathit{l}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathit{P}\mathit{C}{\mathit{l}}_{\mathbf{5}}\left(g\right)$

In a balanced chemical equation, the number of atoms on the reactant side should be same with the number of atoms on the product side for a particular atom. For balancing the reaction,the stochiometric coefficient is multiplied with the P and Cl atoms.

So, the balanced chemical equation is:

${\mathit{P}}_{\mathbf{4}}\left(s\right)\mathbf{+}\mathbf{10}\mathit{C}{\mathit{l}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{4}\mathit{P}\mathit{C}{\mathit{I}}_{\mathbf{5}}\left(g\right)$

The number of moles is calculated by the mass and Molar mass. The relationship between the number of moles, mass, and molar mass is given below.

$\mathit{N}\mathit{u}\mathit{m}\mathit{b}\mathit{e}\mathit{r}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{=}\frac{\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}}{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{}\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}}$

The mass of ${\mathit{P}}_{\mathbf{4}}$= 455 g.

The molar mass of ${\mathit{P}}_{\mathbf{4}}$= 123.895 g/mol.

The number of moles of ${\mathit{P}}_{\mathbf{4}}$is calculated as:

$$\begin{gathered} molesof{\mathit{P}}_{\mathbf{4}}\mathbf{=}\frac{\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}{\mathbf{P}}_{\mathbf{4}}}{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{}\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}{\mathbf{P}}_{\mathbf{4}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{455}\mathbf{g}}{\mathbf{123}\mathbf{.}\mathbf{895}\mathbf{g}\mathbf{/}\mathbf{m}\mathbf{o}\mathbf{l}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{6725}\mathbf{}\mathit{m}\mathit{o}\mathit{l} \end{gathered}$$

Therefore, the number of moles of ${\mathit{P}}_{\mathbf{4}}$is 3.6725 mol.

In the given reaction, 1 mol of ${\mathit{P}}_{\mathbf{4}}$is reacted with 10 mol of$\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathbf{.}$

Thus,

$\mathbf{1}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{o}\mathit{f}\mathbf{}{\mathit{P}}_{\mathbf{4}}\mathbf{=}\mathbf{10}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{C}{\mathit{l}}_{\mathbf{2}}$

The number of moles of $\mathit{C}{\mathit{l}}_{\mathbf{2}}$is:

role="math" localid="1656594065461" $$\begin{gathered} \mathbf{1}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{o}\mathit{f}\mathbf{}{\mathit{P}}_{\mathbf{4}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{10}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{C}\mathit{l} \\ \mathbf{3}\mathbf{.}\mathbf{6725}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{o}\mathit{f}\mathbf{}{\mathit{P}}_{\mathbf{4}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{6725}\mathbf{\times }\mathbf{10}\mathit{m}\mathit{o}\mathit{l}\mathit{o}\mathit{f}\mathbf{}\mathit{C}{\mathit{l}}_{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{36}\mathbf{.}\mathbf{725}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{C}{\mathit{l}}_{\mathbf{2}} \end{gathered}$$

The molar mass of $\mathit{C}{\mathit{l}}_{\mathbf{2}}$= 70.906 g/mol.

The mass of $\mathit{C}{\mathit{l}}_{\mathbf{2}}$is calculated as:

role="math" localid="1656594228465" $$\begin{gathered} \mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathbf{\times }\mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathbf{}\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{C}{\mathit{l}}_{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{36}\mathbf{.}\mathbf{725}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{\times }\mathbf{70}\mathbf{.}\mathbf{906}\mathbf{}\mathit{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{2604}\mathbf{.}\mathbf{023}\mathit{g} \end{gathered}$$

Therefore, the mass of $\mathit{C}{\mathit{l}}_{\mathbf{2}}$is approximately$2.6\times {10}^{3}g$