Chapter 11, Problem 11.56P

a)

Interpretation Introduction

Interpretation:

Balanced chemical equation for reaction of cadmium chloride with silver perchlorate has to be determined.

Concept Introduction:

The reaction that has equal number of atoms of different elements in reactant as well as in product side is called balanced chemical reaction. The chemical equation is balanced to follow the Law of the conservation of mass.

Explanation of Solution

The steps to balance a chemical reaction are as follows:

Step 1: Write the unbalanced equation.

  ${\text{CdCl}}_{\text{2}}+{\text{AgClO}}_{\text{4}}\to \text{AgCl}+\text{Cd}{\left({\text{ClO}}_{\text{4}}\right)}_2$

Step 2:Then write the number of atoms of all elements that are present in chemical reaction in the reactant side and product side.

i) On reactant side,

Number of silver atom is 1.

Number of cadmium atom is 1.

Number of oxygen atoms is 4.

Number of chlorine atom is 3.

ii) On product side,

Number of silver atom is 1.

Number of cadmium atom is 1.

Number of oxygen atoms is 8.

Number of chlorine atom is 3.

Step 3: Balance the number of other atoms of elements except oxygen. The other atom are balanced on both side. Now the reaction is,

  ${\text{CdCl}}_{\text{2}}+{\text{AgClO}}_{\text{4}}\to \text{AgCl}+\text{Cd}{\left({\text{ClO}}_{\text{4}}\right)}_2$

Step 4: After this, balance the number of oxygen atoms. The oxygen atoms are unbalanced on both sides. Multiply ${\text{AgClO}}_{\text{4}}$ by 2 to balance oxygen atom and $\text{AgCl}$ by 2 to balance silver atom as silver atom will become unbalanced. Now the reaction is,

  ${\text{CdCl}}_{\text{2}}+\boxed{2}{\text{AgClO}}_{\text{4}}\to \boxed{2}\text{AgCl}+{\text{CdClO}}_{\text{4}}$

Step 5: Finally, check the number of atoms of each element on both sides. If the number is same then chemical equation is balanced. The balanced chemical equation is,

  ${\text{CdCl}}_{\text{2}}+2{\text{AgClO}}_{\text{4}}\to 2\text{AgCl}+\text{Cd}{\left({\text{ClO}}_{\text{4}}\right)}_2$

b)

Interpretation Introduction

Interpretation:

Mass of $\text{AgCl}$ produced from the reaction has to be determined.

Concept Introduction:

Mole is S.I. unit. The number of moles is calculated as ratio of mass of compound to molar mass of compound.

Molar mass is sum of the total mass in grams of all atoms that make up mole of particular molecule that is mass of 1 mole of compound. The S.I unit is $\text{g/mol}$.

The expression to relate number of moles, mass and molar mass of compound is as follows:

  $\text{Number of moles}=\frac{\text{mass of the compound}}{\text{molar mass of the compound}}$

For more than one quantity of reactants in a reaction, the reactant that is completely consumed and controls the amount of product synthesized is called limiting reactant. The other reactants are called excess reactants. Also, the limiting reactant is completely consumed the initial mass of limiting reactant is used to calculate mass of product.

Answer to Problem 11.56P

Mass of $\text{AgCl}$ produced from the reaction is $11.4710\text{8 g}$.

Explanation of Solution

The reaction is as follows:

    ${\text{CdCl}}_{\text{2}}+2{\text{AgClO}}_{\text{4}}\to 2\text{AgCl}+\text{Cd}{\left({\text{ClO}}_{\text{4}}\right)}_2$

The formula to calculate moles of ${\text{CdCl}}_{\text{2}}$ is as follows:

  ${\text{Number of moles of CdCl}}_{\text{2}}=\frac{{\text{mass of CdCl}}_{\text{2}}}{{\text{molar mass of CdCl}}_{\text{2}}}$        (1)

Substitute $17.5\text{ g}$ for mass of ${\text{CdCl}}_{\text{2}}$ and $183.317\text{ g/mol}$ for molar mass of ${\text{CdCl}}_{\text{2}}$ in equation (1).

  $\begin{array}{c}{\text{Number of moles of CdCl}}_{\text{2}}=\left(\frac{17.5\text{ g}}{\text{183}\text{.317 g/mol}}\right)\\ =0.09546\text{ mol}\end{array}$

The formula to calculate moles of ${\text{AgClO}}_{\text{4}}$ is as follows:

  $\text{Number of moles of}{\text{AgClO}}_{\text{4}}=\frac{\text{mass of}{\text{AgClO}}_{\text{4}}}{{\text{molar massof AgClO}}_{\text{4}}}$        (2)

Substitute $35.5\text{ g}$ for mass of ${\text{AgClO}}_{\text{4}}$ and $207.3188\text{ g/mol}$ for molar mass of ${\text{AgClO}}_{\text{4}}$ in equation (2).

  $\begin{array}{c}\text{Number of moles of}{\text{AgClO}}_{\text{4}}=\left(\frac{35.5\text{ g}}{207.3188\text{ g/mol}}\right)\\ =0.1712\text{ mol}\end{array}$

To find the limiting reactant, divide the number of moles of each reactant by their stoichiometric coefficients.

i) The stoichiometric coefficient of ${\text{CdCl}}_{\text{2}}$ is 1 and number of moles of ${\text{CdCl}}_{\text{2}}$ is $0.09546\text{ mol}$. Therefore, number of molesof ${\text{CdCl}}_{\text{2}}$ will be divided by 1. Hence, number of moles ${\text{CdCl}}_{\text{2}}$ will be $0.09546\text{ mol}$.

ii)The stoichiometric coefficient of ${\text{AgClO}}_{\text{4}}$ is 2 and number of moles of ${\text{AgClO}}_{\text{4}}$ is $0.1712\text{ mol}$. Therefore, number of moles of ${\text{AgClO}}_{\text{4}}$ will be divided by 2. Hence, number of moles ${\text{AgClO}}_{\text{4}}$ will be $0.0856\text{ mol}$.

The number of moles of ${\text{AgClO}}_{\text{4}}$ is less in comparison to other reactants in reaction. Hence, ${\text{AgClO}}_{\text{4}}$ is limiting reagent. Also, initial mass of ${\text{AgClO}}_{\text{4}}$ is used to calculate mass of product formed.

The reaction between ${\text{CdCl}}_{\text{2}}$ and ${\text{AgClO}}_{\text{4}}$ is as follows:

    ${\text{CdCl}}_{\text{2}}+2{\text{AgClO}}_{\text{4}}\to 2\text{AgCl}+\text{Cd}{\left({\text{ClO}}_{\text{4}}\right)}_2$

According to stoichiometry of reaction, 2 moles of ${\text{AgClO}}_{\text{4}}$ produces 2 moles of $\text{AgCl}$. Therefore, number of moles of $\text{AgCl}$ is equal to number of moles of ${\text{AgClO}}_{\text{4}}$ and that is $0.1712\text{ mol}$.

The formula to calculate mass of $\text{AgCl}$ is as follows:

    $\text{Mass of AgCl}=\left(\text{number of moles}\right)\left(\text{molar mass}\right)$        (3)

Substitute $\text{143}\text{.3212 g/mol}$ for molar mass of $\text{AgCl}$ and $0.1712\text{ mol}$ for number of moles of $\text{AgCl}$ in equation (3).

  $\begin{array}{c}\text{Mass of AgCl}=\left(0.1712\text{ mol}\right)\left(\text{143}\text{.3212 g/mol}\right)\\ =24.5365\text{ g}\end{array}$

c)

Interpretation Introduction

Interpretation:

Mass of excess reactant among ${\text{CdCl}}_{\text{2}}$ and ${\text{AgClO}}_{\text{4}}$ has to be determined.

Concept Introduction:

Refer to part (b).

Answer to Problem 11.56P

Mass of excess ${\text{CdCl}}_{\text{2}}$ is $1.8075\text{ g}$.

Explanation of Solution

The reaction between ${\text{CdCl}}_{\text{2}}$ and ${\text{AgClO}}_{\text{4}}$ is as follows:

    ${\text{CdCl}}_{\text{2}}+2{\text{AgClO}}_{\text{4}}\to 2\text{AgCl}+\text{Cd}{\left({\text{ClO}}_{\text{4}}\right)}_2$

The number of moles of ${\text{AgClO}}_{\text{4}}$ is less in comparison to other reactants in reaction. Hence, ${\text{AgClO}}_{\text{4}}$ is limiting reagent and ${\text{CdCl}}_{\text{2}}$ will be excess reagent.

According to stoichiometry of reaction, 2 moles of ${\text{AgClO}}_{\text{4}}$ reacts with 1 mole of ${\text{CdCl}}_{\text{2}}$. Therefore, number of moles of ${\text{CdCl}}_{\text{2}}$ is 1/2 times number of moles of ${\text{AgClO}}_{\text{4}}$.

The formula to calculate number of moles of ${\text{CdCl}}_{\text{2}}$ is as follows:

    ${\text{Number of moles of CdCl}}_{\text{2}}=\frac{1}{2}\left({\text{Number of moles of AgClO}}_4\right)$        (4)

Substitute $0.1712\text{ mol}$ for number moles of ${\text{AgClO}}_{\text{4}}$ in equation (4).

  $\begin{array}{c}{\text{Number of moles of CdCl}}_{\text{2}}=\frac{1}{2}\left(0.1712\text{ mol}\right)\\ =0.08\text{56mol}\end{array}$

Excess mole of ${\text{CdCl}}_{\text{2}}$ is calculated as follows:

    ${\text{Excess mole of CdCl}}_{\text{2}}={\text{moles of CdCl}}_{\text{2}}-{\text{mole of CdCl}}_{\text{2}}\text{ required}$        (5)

Substitute, $0.09546\text{ mol}$ for moles of ${\text{CdCl}}_{\text{2}}$ and $0.0856\text{ mol}$ for moles of ${\text{CdCl}}_{\text{2}}$ require in equation (5).

  $\begin{array}{c}{\text{Excess mole of CdCl}}_{\text{2}}=0.09546\text{ mol}-0.0856\text{ mol}\\ =0.0098\text{6 mol}\end{array}$

The formula to calculate mass of ${\text{CdCl}}_{\text{2}}$ in excess is as follows:

    ${\text{Mass of excess of CdCl}}_{\text{2}}=\left(\text{number of moles in excess}\right)\left(\text{molar mass}\right)$        (6)

Substitute $0.0098\text{6 mol}$ for excess moles of ${\text{CdCl}}_{\text{2}}$ and $183.317\text{ g/mol}$ for molar mass of ${\text{CdCl}}_{\text{2}}$ in equation (6).

  $\begin{array}{c}{\text{Mass of excess of CdCl}}_{\text{2}}=\left(0.00986\text{ mol}\right)\left(183.317\text{ g/mol}\right)\\ =1.8075\text{ g}\end{array}$