Question

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?

Short answer

Answer

The mass of ${\mathbf{Cl}}_{\mathbf{2}}$is$\mathbf{2.6}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{g}\mathbf{.}$

Step-by-step solution

Chemical equation of the reaction of P4 with Cl2

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:

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

Balance the chemical equation

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:

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

Relation between mass and number of moles

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.

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

Calculate the number of moles of P4

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.

Relation between number of moles of P4 and Cl2

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,

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

Calculate the number of moles of Cl2

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

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

Calculate the mass of 

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$2.6\times {10}^{3}g$