Chapter 9, Problem 18A

Interpretation Introduction

Interpretation:

The mass of water needed to completely react with 20.0 g of PCl₃ needs to be deduced based on the given reaction.

Concept Introduction:

• A chemical reaction is expressed as a chemical equation having reactants and products on the left and right side of the reaction arrow respectively.

  $\text{Reactants }\to \text{ Products}$

• The coefficient of a balanced chemical equation, i.e., the stoichiometry gives the number of reactants and products involved in the reaction.
• The stoichiometric ratio between the reactants or between reactants and products is expressed in terms of the mole-mole ratio.

Answer to Problem 18A

Mass of water = 2.62 g

Explanation of Solution

The given reaction is:

  ${\text{PCl}}_3{\text{(s) + 3H}}_{\text{2}}\text{O(l) }\to {\text{H}}_3{\text{PO}}_3\text{(aq) + 3HCl(aq) }$

Step 1: Calculate the moles of PCl₃ present as follows:

Mass of PCl₃ present = 20.0 g

Molecular weight of PCl₃ = 137.33 g/mole

The moles will be:

  $\begin{array}{c}{\text{Moles of PCl}}_{\text{3}}\text{ = }\frac{{\text{Mass of PCl}}_{\text{3}}}{{\text{Molar Mass PCl}}_{\text{3}}}\end{array}$

         $\begin{array}{c}=\frac{20.0\text{ g}}{137.33\text{ g/mol}}\end{array}$

         $\begin{array}{c}=0.1456\text{ moles}\end{array}$

Step 2: Calculate the moles of water reacted:

Based on the reaction stoichiometry:

1 mole of PCl₃ reacts with 3 moles of water.

Therefore, 0.1456 moles of PCl₃ would react with:

  $\frac{0.1456{\text{ moles PCl}}_{\text{3}}\times {\text{ 3 moles H}}_{\text{2}}\text{O}}{1{\text{ mole PCl}}_{\text{3}}}=0.145{\text{6 moles H}}_{\text{2}}\text{O}$

Step 3: Calculate the mass of water needed as follows:

Moles of water reacted = 0.1456 moles

Molecular weight of H₂O = 18 g/mol

  $\begin{array}{c}{\text{Mass of H}}_{\text{2}}\text{O = Moles }\times \text{ Molecular weight}\end{array}$

        $\begin{array}{c}=0.1456\text{ moles }\times \text{ 18 g/mol }\end{array}$

        $\begin{array}{c}\text{= 2}\text{.62 g}\end{array}$

Conclusion

Therefore, 2.62 g of water is needed to completely react with 20.0 g of PCl₃.