Chapter 11, Problem 81A

(a)

Interpretation Introduction

Interpretation:

Limiting reactant is to be calculated if 25 g of Zn and 30 g of MnO₂ are reacted.

Concept introduction:

1 mol is defined as $6.02214076\times {10}^{23}$ particles of any substance.

Molecular mass is defined as the mass of one mole of substance.

The formula to calculate moles of substance

$\text{Moles=}\frac{\begin{array}{ccc}\text{Mass}& \text{of}& \text{substance}\end{array}}{\begin{array}{cccc}\text{Molar}& \text{mass}& \text{of}& \text{substance}\end{array}}$

Answer to Problem 81A

MnO₂ act as the limiting reagent and Zn is in excess

Explanation of Solution

The given compound is MnO₂.

The formula to calculate formula mass is,

$\text{Formula}\text{mass}=\left(N_1\times x_1\right)+\left(N_2\times x_2\right)+\mathrm{...}$

Where,

• $N_1$ represents the number of atoms of first element in chemical formula.
• ${\text{x}}_1$ represents the atomic mass of first element.
• $N_2$ represents the number of atoms of second element in chemical formula.
• ${\text{x}}_1$ represents the atomic mass of second element.

The atomic mass of Manganese is 54.938 g / mol.

The atomic mass of oxygen is $15.99\text{g}/\text{mol}$.

Substitute the value of atomic mass in the above formula.

$\begin{array}{c}\text{Formula}\text{mass}=\left(N_1\times x_1\right)+\left(N_2\times x_2\right)+\mathrm{...}\\ =\left(1\times 54.938\right)+\left(2\times 15.99\right)\\ =86.9368\text{g/mol}\end{array}$

Molecular mass of MnO₂ is 86.9368 g/mol.

The given compound is Zn(OH)₂.

The formula to calculate formula mass is,

$\text{Formula}\text{mass}=\left(N_1\times x_1\right)+\left(N_2\times x_2\right)+\mathrm{...}$

Where,

• $N_1$ represents the number of atoms of first element in chemical formula.
• ${\text{x}}_1$ represents the atomic mass of first element.
• $N_2$ represents the number of atoms of second element in chemical formula.
• ${\text{x}}_1$ represents the atomic mass of second element.

The atomic mass of Zinc is 65.38 g / mol.

The atomic mass of oxygen is $15.99\text{g}/\text{mol}$.

The atomic mass of hydrogen is $1.00\text{g}/\text{mol}$.

Substitute the value of atomic mass in the above formula.

$\begin{array}{c}\text{Formula}\text{mass}=\left(N_1\times x_1\right)+\left(N_2\times x_2\right)+\mathrm{...}\\ =\left(1\times 65.38\right)+\left(2\times 15.99\right)+(2\times 1.0)\\ =99.394\text{g/mol}\end{array}$

Molecular mass of Zn(OH)₂ is 99.394 g/mol.

$\begin{array}{c}numberofmolesofZn=\frac{MassofZn}{MolarmassofZn}\\ =\frac{25g}{65.38g/mol}\\ =0.3823moles\end{array}$

$\begin{array}{c}numberofmolesofMnO2=\frac{MassofMnO2}{MolarmassofMnO2}\\ =\frac{30g}{86.93g/mol}\\ =0.3451moles\end{array}$

In alkaline battery electrical energy is produced according to below chemical reaction.

$Zn+2MnO2+H2O\to Zn(OH)2+Mn2O3$

From the stoichiometry of reaction.

For Zinc

$\begin{array}{l}1molesofZnproduce=1molesofZn(OH)2\\ 0.3823\times 1molesofZnproduce=0.3823\times 1molesofZn(OH)2\\ 0.3823molesofZnproduce=0.3823molesofZn(OH)2\end{array}$

For MnO₂

$\begin{array}{l}2molesofMnO2produce=1molesofZn(OH)2\\ 1molesofMnO2produce=\frac{1}{2}molesofZn(OH)2\\ 0.3451\times 1molesofMnO2produce=0.3451\times \frac{1}{2}molesofZn(OH)2\\ 0.3451molesofMnO2produce=0.1725molesofZn(OH)2\end{array}$

The reactant which produce least amount of product is limiting reagent. So, MnO₂ act as the limiting reagent and Zn is in excess.

(b)

Interpretation Introduction

Interpretation:

Mass of Zn(OH)₂ produced to be calculated.

Concept introduction:

1 mol is defined as $6.02214076\times {10}^{23}$ particles of any substance.

Molecular mass is defined as the mass of one mole of substance.

The formula to calculate moles of substance

$\text{Moles=}\frac{\begin{array}{ccc}\text{Mass}& \text{of}& \text{substance}\end{array}}{\begin{array}{cccc}\text{Molar}& \text{mass}& \text{of}& \text{substance}\end{array}}$

Answer to Problem 81A

17.145 g of Zn(OH)₂ formed during the reaction.

Explanation of Solution

Moles of Zn(OH)₂ was calculated above.

Number of moles of Zn(OH)₂ produced by limiting reagent is the actual amount of Zn(OH)₂

formed and it is equal to 0.1725.

$\begin{array}{c}Massofsubs\tan ce=moles\times molarmassofsubs\tan ce\\ =0.1725\times 99.394\\ =17.145g\end{array}$