Chapter 13.3, Problem 42PP

Interpretation Introduction

Interpretation:

The volume of oxygen required to produce $34L$ dinitrogen oxide is to be determined.

Concept introduction:

It involves stochiometry concerns and Avogadro’s law.

Answer to Problem 42PP

The volume of oxygen required is $17L$

Explanation of Solution

The given equation is:

$N{H}_{4}N{O}_3\to N_{2}O+2H_{2}O$

Equation shows $1mol$

$N{H}_{4}N{O}_3$ forms $1mol{N}_{2}O$

Volume of N₂O is 0.1 L.

Therefore $0.1mol$

$N{H}_{4}N{O}_3$ will form $0.1mol{N}_{2}O$

Applying ideal gas equation:

$\mathrm{PV}=nRT$

$\mathrm{n}=\frac{PV}{RT}$

Where

$\mathrm{P}$ is pressure of the gas

$\mathrm{V}$ is volume of the gas

$\mathrm{n}$ is the number of moles of the gas

$\mathrm{R}$ is the ideal gas constant

$\mathrm{T}$ is the temperature of the gas

Further, mass of gas $\mathrm{m}$ is calculated by multiplying number of moles $\mathrm{n}$ with molar mass of the gas $\mathrm{M}$

Given data is:

$\text{V=0}.1L$

At STP

$\text{P=}1atm$

$\text{T=}273K$

$\text{R=}0.0821Latm{K}^{-1}mo{l}^{-1}$

On substituting the values in the formula:

$\begin{array}{c}\mathrm{n}=\frac{1\times 0.1}{\begin{array}{l}0.0821\times 273\\ \end{array}}\\ =0.004\text{mol}\end{array}$

So $\text{0}\text{.004}\text{mol}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{is}\text{produced}\text{by}\text{0}\text{.1}\text{L}{\text{N}}_{\text{2}}\text{O}$

$\mathrm{m}=M\times n$

The given compound is ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$.

The formula to calculate molar mass is,

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

Where,

${\mathrm{N}}_1$

• represents the number of atoms of first element in chemical formula.
• ${\mathrm{x}}_1$
• represents the atomic mass of first element.
• ${\mathrm{N}}_2$
• represents the number of atoms of second element in chemical formula.
• ${\text{x}}_1$
• represents the atomic mass of second element.

Atomic mass of nitrogen is $14gmo{l}^{-1}$

Atomic mass of oxygen is $16gmo{l}^{-1}$

Atomic mass of hydrogen is $1gmo{l}^{-1}$

Hence, $\begin{array}{c}M=14+1\times 4+14+3\times 16\\ =80\end{array}$

$$

$\begin{array}{c}\mathrm{m}=80\times 0.004\\ =0.32g\end{array}$

Therefore mass is $0.32g$

Conclusion

$0.32gN{H}_{4}N{O}_3$ is produced by $0.1L{N}_{2}O$