Chapter 12, Problem 53A

a.

Interpretation Introduction

Interpretation: To identify the limiting reagent for the reaction.

Concept Introduction: The formation of products is dependent on the number of reactants. The reactant that is consumed first is known as the limiting reactant, when this reactant is consumed completely, then there will be no more formation of products.

Answer to Problem 53A

Aluminum is the limiting reagent.

Explanation of Solution

The given chemical reaction is depicted as follows:

  $2\text{ Al}\left(s\right)+3{\text{Cl}}_2\left(g\right)\to 2{\text{ AlCl}}_3\left(s\right)$

When two moles of aluminum react with three moles of chlorine, it yields two moles of aluminum chloride.

Calculation for moles of chlorine is as follows:

  $3.0\cancel{\text{mol Al}}\times \frac{3{\text{mol Cl}}_2}{2\cancel{\text{mol Al}}}=4.5{\text{ mol Cl}}_2$

Calculation for moles of aluminum is as follows:

  $5.3\cancel{{\text{mol Cl}}_2}\times \frac{2\text{ mol Al}}{3\cancel{{\text{mol Cl}}_2}}=3.22\text{ mol Al}$

Thus, the reactant that is consumed first is Aluminum as the number of moles is less than chlorine.

b.

Interpretation Introduction

Interpretation: To calculate the number of moles of product formed.

Concept Introduction: A mole is $6.02\times {10}^{23}$ atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The coefficients are the numbers that are written in front of the chemical formula/symbol in order to balance a chemical reaction. The coefficients help in determining the conversion factors between the number of moles for two different substances in a chemical equation.

Answer to Problem 53A

The number of products are $3\text{ moles}$ .

Explanation of Solution

The given chemical reaction is depicted as follows:

  $2\text{ Al}\left(s\right)+3{\text{Cl}}_2\left(g\right)\to 2{\text{ AlCl}}_3\left(s\right)$

When two moles of aluminum react with three moles of chlorine, it yields two moles of aluminum chloride.

Calculation for moles of aluminum chloride is as follows:

  $3.0\cancel{\text{mol Al}}\times \frac{{\text{2 mol AlCl}}_3}{2\cancel{\text{mol Al}}}=3.0{\text{ mol AlCl}}_3$

c.

Interpretation Introduction

Interpretation: To calculate the number of moles of excess reagent.

Concept Introduction: A mole is $6.02\times {10}^{23}$ atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The coefficients are the numbers that are written in front of the chemical formula/symbol in order to balance a chemical reaction. The coefficients help in determining the conversion factors between the number of moles for two different substances in a chemical equation.

Answer to Problem 53A

The number of moles that is present in excess is $\text{0}\text{.8 mol}$ .

Explanation of Solution

The given chemical reaction is depicted as follows:

  $2\text{ Al}\left(s\right)+3{\text{Cl}}_2\left(g\right)\to 2{\text{ AlCl}}_3\left(s\right)$

When two moles of aluminum react with three moles of chlorine, it yields two moles of aluminum chloride.

Since aluminum is the limiting reactant in this reaction. When the reaction gets stops, chlorine is the reactant that is still present in the reaction.

The moles of chlorine before the reaction is $\text{5}\text{.3 mol}$ .

Calculation for moles of chlorine is as follows:

  $3.0\cancel{\text{mol A}}\text{l}\times \frac{3{\text{mol Cl}}_2}{2\cancel{\text{mol Al}}}=4.5{\text{ mol Cl}}_2$

The amount of chlorine that is still present in the reaction is calculated as follows:

  $\begin{array}{l}\text{5}\text{.3 mol}-4.5\text{ mol}\\ =\text{0}\text{.8 mol}\end{array}$