Chapter 11, Problem 80A

Interpretation Introduction

Interpretation:

The limiting reactant and excess reactant needs to be identified.

Concept introduction:

1 mol is defined as $6.02214076\times {10}^{23}$ particles of any substance.

Molecular mass is defined as the mass of one mole of substance.

The formula to calculate moles of substance

$\text{Moles=}\frac{\begin{array}{ccc}\text{Mass}& \text{of}& \text{substance}\end{array}}{\begin{array}{cccc}\text{Molar}& \text{mass}& \text{of}& \text{substance}\end{array}}$

Answer to Problem 80A

Cl₂ act as the limiting reagent and P₄ is in excess.

Explanation of Solution

$\text{10Cl2+P4}\to \text{4PCl5}$

Chlorine and Phosphorus reacts according to the above reaction.

The given compound is PCl₅.

The formula to calculate formula mass is,

$\text{Formula}\text{mass}=\left(N_1\times x_1\right)+\left(N_2\times x_2\right)+\mathrm{...}$

Where,

• $N_1$ represents the number of atoms of first element in chemical formula.
• ${\text{x}}_1$ represents the atomic mass of first element.
• $N_2$ represents the number of atoms of second element in chemical formula.
• ${\text{x}}_1$ represents the atomic mass of second element.

The atomic mass of Phosphorusis 30.97g / mol.

The atomic mass of chlorine is $35.5\text{g}/\text{mol}$.

Substitute the value of atomic mass in the above formula.

$\begin{array}{c}\text{Formula}\text{mass}=\left(N_1\times x_1\right)+\left(N_2\times x_2\right)+\mathrm{...}\\ =\left(1\times 30.97\right)+\left(5\times 35.5\right)\\ =208.23\text{g/mol}\end{array}$

The formula mass or molar mass of PCl₅ is $208.23\text{g/mol}$.

Molar Mass of Cl₂ is 70.90 g/mol

Molar Mass of P₄ is 123.89 g/mol

$\begin{array}{c}numberofmolesofCl2=\frac{MassofCl2}{MolarmassofCl2}\\ =\frac{16g}{70.90g/mol}\\ =0.2256moles\end{array}$

$\begin{array}{c}numberofmolesofP4=\frac{MassofP4}{MolarmassofP4}\\ =\frac{23g}{123.89g/mol}\\ =0.1856moles\end{array}$

According to stoichiometry of reaction

For Cl₂

$\begin{array}{l}\text{1}\text{0}\text{moles}\text{of}\text{Cl2}\text{}\text{produce}\text{=}\text{4}\text{moles}\text{of}\text{PCl5}\\ \text{1}\text{moles}\text{of}\text{Cl2}\text{}\text{produce}\text{=}\frac{\text{4}}{\text{10}}\text{moles}\text{of}\text{PCl5}\\ \text{0}\text{.2256×1}\text{moles}\text{of}\text{Cl2}\text{}\text{produce}\text{=}\text{0}\text{.2256×}\frac{\text{4}}{\text{10}}\text{moles}\text{of}\text{PCl5}\\ \text{0}\text{.2256}\text{moles}\text{of}\text{Cl2}\text{}\text{produce}\text{=}\text{0}\text{.09024}\text{moles}\text{of}\text{PCl5}\end{array}$

For P₄

$\begin{array}{l}\text{1}\text{moles}\text{of}\text{P4}\text{}\text{produce}\text{=}\text{4}\text{moles}\text{of}\text{PCl5}\\ \text{0}\text{.1856×1}\text{moles}\text{of}\text{P4}\text{}\text{}\text{produce}\text{=}\text{0}\text{.1856×4}\text{moles}\text{of}\text{PCl5}\\ \text{0}\text{.1856}\text{moles}\text{of}\text{P4}\text{}\text{}\text{produce}\text{=}\text{0}\text{.7424}\text{moles}\text{of}\text{PCl5}\end{array}$

The reactant which produce least amount of product is limiting reagent. So, Cl₂ act as the limiting reagent and P₄ is in excess.

Conclusion

Cl₂ act as the limiting reagent and P₄ is in excess.