Chapter 9, Problem 61A

Interpretation Introduction

Interpretation:

The maximum mass of phosphoric acid formed needs to be determined.

  $P_4\text{}(s)+O_2\text{}(g)\to P_4\text{}{O}_{10}\text{}(s)$

  $P_4\text{}{O}_{10}\text{}(s)+H_{2\text{}}O(l)\to H_{3\text{}}P{O}_4\text{}(l)$

The starting mass is 20.0 g of phosphorus, 30 g of oxygen gas, and 15.0 g of water.

Concept Introduction:

The theoretical yield is the maximum amount of product formed in the reaction. The amount of product formed depends on the amount of limiting reactant given in the reaction.

Answer to Problem 61A

The limiting reagent is H₂ O and the theoretical yield of H₃ PO₄ is 54.4g.

This theoretical yield will be the maximum mass of H₃ PO₄ formed.

Explanation of Solution

The reaction is as follows:

  $P_4\text{}(s)+O_2\text{}(g)\to P_4\text{}{O}_{10}\text{}(s)$

First step is to determine the molar mass of all reactants in the reaction as;

  $\begin{array}{l}M,P_{4\text{}}=(4P\times 31)=124g/mol\\ M,O_2\text{}=(2O\times 16)=32g/mol\\ M,P_4\text{}{O}_{10}\text{}=(4P\times 31)+(10O\times 16)=284g/mol\end{array}$

The balanced equation can be expressed as:

  $P_4\text{}(s)+5O_2\text{}(g)\to P_4\text{}{O}_{10}\text{}(s)$

The theoretical yield of P₄ O₁₀ is determined by the limiting reagent and the excess reagent

  $\text{20}{\text{.0gP}}_{\text{4}}\text{×}\frac{{\text{1molP}}_{\text{4}}}{{\text{124gP}}_{\text{4}}}\text{×}\frac{{\text{1molP}}_{\text{4}}{\text{O}}_{\text{10}}}{{\text{1molP}}_{\text{4}}}\text{×}\frac{{\text{284gP}}_{\text{4}}{\text{O}}_{\text{10}}}{{\text{1molP}}_{\text{4}}{\text{O}}_{\text{10}}}\text{=45}{\text{.8gP}}_{\text{4}}{\text{O}}_{\text{10}}$

And,

  $\text{30}{\text{.0gP}}_{\text{4}}\text{×}\frac{{\text{1molO}}_{\text{2}}}{{\text{30gO}}_{\text{2}}}\text{×}\frac{{\text{1molP}}_{\text{4}}{\text{O}}_{\text{10}}}{{\text{1molO}}_{\text{2}}}\text{×}\frac{{\text{284gP}}_{\text{4}}{\text{O}}_{\text{10}}}{{\text{1molP}}_{\text{4}}{\text{O}}_{\text{10}}}\text{=53}{\text{.3gP}}_{\text{4}}{\text{O}}_{\text{10}}$

The theoretical yield is the one that gives in the smallest proportion and that is determined with the limiting reagent.

The limiting reagent is P₄ and the theoretical yield of P₄ O₁₀ is 45.8 g.

Now,

  $P_4\text{}{O}_{10}\text{}(s)+H_{2\text{}}O(l)\to H_{3\text{}}P{O}_4\text{}(l)$

Determine the molar mass of all reactants in the reaction

  $\begin{array}{l}{\text{M,H}}_{\text{2}\text{}}\text{O=(2H×1)+(1O×16)=18g/mol}\\ {\text{M,H}}_{\text{3}\text{}}{\text{PO}}_{\text{4}\text{}}\text{=(3H×1)+(1P×31)+(4O×16)=98g/mol}\end{array}$

The equation is balanced as follows:

  $P_4\text{}{O}_{10}\text{}(s)+6H_{2\text{}}O(l)\to 4H_{3\text{}}P{O}_4\text{}(l)$

The theoretical yield of P₄O₁₀is determined by the limiting reagent and the excess reagent as follows:

  $\begin{array}{l}\text{45}{\text{.8gP}}_{\text{4}}{\text{O}}_{\text{10}}\text{×}\frac{{\text{1molP}}_{\text{4}}{\text{O}}_{\text{10}}}{{\text{124gP}}_{\text{4}}{\text{O}}_{\text{10}}}\text{×}\frac{{\text{1molH}}_{\text{3}}{\text{PO}}_{\text{4}}}{{\text{1molP}}_{\text{4}}{\text{O}}_{\text{10}}}\text{×}\frac{{\text{98gH}}_{\text{3}}{\text{PO}}_{\text{4}}}{{\text{1molH}}_{\text{3}}{\text{PO}}_{\text{4}}}\text{=63}{\text{.2gH}}_{\text{3}}{\text{PO}}_{\text{4}}\\ \text{15}{\text{.0gH}}_{\text{2}}\text{O×}\frac{{\text{1molH}}_{\text{2}}\text{O}}{{\text{18gH}}_{\text{2}}\text{O}}\text{×}\frac{{\text{4molH}}_{\text{3}}{\text{PO}}_{\text{4}}}{{\text{6molH}}_{\text{2}}\text{O}}\text{×}\frac{{\text{98gH}}_{\text{3}}{\text{PO}}_{\text{4}}}{{\text{1molH}}_{\text{3}}{\text{PO}}_{\text{4}}}\text{=54}{\text{.4gH}}_{\text{3}}{\text{PO}}_{\text{4}}\end{array}$

The limiting reagent is H₂ O and the theoretical yield of H₃ PO₄ is 54.4g.

This theoretical yield is the maximum mass formed.