Chapter 9, Problem 65A

Interpretation Introduction

Interpretation: The mass of phosphoric acid that can be formed for the given reaction needs to be determined.

  $\begin{array}{l}{\text{P}}_{\text{4(s) }}{\text{+ O}}_{\text{2(g) }}\stackrel{}{\to }{\text{P}}_{\text{4}}{\text{O}}_{\text{10(s)}}\\ {\text{P}}_{\text{4}}{\text{O}}_{\text{10(s)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{PO}}_{\text{4(l)}}\end{array}$

Concept Introduction:A chemical reaction involves the conversion of reactant to product by breaking and making of chemical bonds.

There can be more than one reactant molecules involve in a chemical reaction to form the products. The reactant which is present in limited amount in a chemical reaction is called as limiting reactant.

Answer to Problem 65A

  $\text{54}{\text{.4 g H}}_{\text{3}}{\text{PO}}_{\text{4}}$

Explanation of Solution

Given information:

Mass of P = 20.0 g

Mass of O₂ = 30.0 g

Mass of water = 15.0 g

The reactions are represented as follows:

  $\begin{array}{l}{\text{P}}_{\text{4(s) }}{\text{+ 5 O}}_{\text{2(g) }}\stackrel{}{\to }{\text{P}}_{\text{4}}{\text{O}}_{\text{10(s)}}\\ {\text{P}}_{\text{4}}{\text{O}}_{\text{10(s)}}{\text{+ 6 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{ 4H}}_{\text{3}}{\text{PO}}_{\text{4(l)}}\end{array}$

Mass of P₄= 20.0 g

Mass of O₂ = 30.0 g

Molar mass of P₄ = 123.8 g/mol

Molar mass of O₂ = 32.0 g/mol

For the reaction (1):

Calculate moles of both reactant and determine the limiting reactant:

  $\begin{array}{l}{\text{Moles of P}}_{\text{4}}\text{= }\frac{\text{20}\text{.00}}{\text{123}\text{.8 g/mol}}\text{= 0}\text{.16 moles}\\ {\text{Moles of O}}_{\text{2}}\text{= }\frac{\text{30}\text{.0 g}}{\text{32}\text{.0 g/mol}}\text{= 0}\text{.937 moles}\end{array}$

  $\text{0}{\text{.16 moles P}}_{\text{4 }}\text{×}\frac{{\text{5 mole O}}_{\text{2}}}{{\text{1 mole P}}_{\text{4 }}}\text{ = 0}{\text{.80 moles O}}_{\text{2}}$

Since the calculated moles of ${\text{O}}_{\text{2}}$ is more therefore ${\text{P}}_{\text{4}}$ must be limiting reactant and will determine the amount of product formed.

  $\text{0}{\text{.16 moles P}}_{\text{4 }}\text{×}\frac{{\text{1 mole P}}_{\text{4}}{\text{O}}_{\text{1}0}}{{\text{1 mole P}}_{\text{4 }}}\text{ = 0}{\text{.16 moles P}}_{\text{4}}{\text{O}}_{\text{1}0}$

Mass of water = 15.0 g

Molar mass of water = 18.0 g/mol

Calculate moles of water:

  ${\text{Moles of H}}_{\text{2}}\text{O= }\frac{\text{15}\text{.0 g}}{\text{18}\text{.0 g/mol}}\text{= 0}\text{.833 moles}$

Determine the limiting reactant for equation (2):

  $\text{0}{\text{.833 moles H}}_{\text{2}}\text{O }\times \frac{{\text{1 mole P}}_{\text{4}}{\text{O}}_{\text{10}}}{{\text{6 moles H}}_{\text{2}}\text{O}}\text{= 0}{\text{.139 mole P}}_{\text{4}}{\text{O}}_{\text{10}}$

Since the calculated moles of P₄O₁₀ are more so H₂O must be limiting reactant.

  $\text{0}{\text{.833 moles H}}_{\text{2}}\text{O }\times \frac{{\text{4 mole H}}_{\text{3}}{\text{PO}}_{\text{4}}}{{\text{6 moles H}}_{\text{2}}\text{O}}\text{= 0}{\text{.555 mole H}}_{\text{3}}{\text{PO}}_{\text{4}}$

Calculate mass of H₃PO₄ as follows:

  $\text{0}{\text{.555moles H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{ ×}\frac{\text{ 97}{\text{.99 g H}}_{\text{3}}{\text{PO}}_{\text{4}}}{{\text{1 mole H}}_{\text{3}}{\text{PO}}_{\text{4}}}\text{= 54}{\text{.4 g H}}_{\text{3}}{\text{PO}}_{\text{4}}$

Conclusion

Thus, mass is $\text{54}{\text{.4 g H}}_{\text{3}}{\text{PO}}_{\text{4}}$