Chapter 3, Problem 3C.15E

(a)

Interpretation Introduction

Interpretation:

Mass of product $\left({\text{NH}}_4\text{Cl}\right)$ produced in reaction has to be determined.

Concept Introduction:

Ideal gas law can be represented as equation for state for any imaginary gas. Expression for ideal gas equation is as follows:

  $PV=nRT$

Here,

$P$ is pressure of the gas.

$V$ is volume of gas.

$n$ denotes moles of a gas.

$R$ is gas constant.

$T$ is temperature of gas.

Answer to Problem 3C.15E

Mass of product $\left({\text{NH}}_4\text{Cl}\right)$ produced in reaction is $4.25\times {10}^{-3}\text{ g}$.

Explanation of Solution

Balance reaction is given as follows:

  ${\text{NH}}_{\text{3}}\left(\text{g}\right)+\text{HCl}\left(\text{g}\right)\to {\text{NH}}_{\text{4}}\text{Cl}\left(\text{g}\right)$

The expression to calculate moles of gas is as follows:

  $n=\frac{PV}{RT}$        (1)

Conversion factor to convert $30\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273.15\\ =30\text{°C}+273.15\\ =303.15\text{K}\end{array}$

Conversion factor to convert $25\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273.15\\ =25\text{°C}+273.15\\ =298.15\text{K}\end{array}$

Substitute $100\text{ Torr}$ for $P$, $15.0\text{ mL}$ for $n$, $62.364\text{ L}\cdot \text{Torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$ and $303.15\text{K}$ for $T$ in equation (1) to calculate moles of ${\text{NH}}_3$ gas.

  $\begin{array}{c}n=\left(\frac{\left(100\text{ Torr}\right)\left(15.0\text{ mL}\right)}{\left(62.364\text{ L}\cdot \text{Torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(303.15\text{K}\right)}\right)\left(\frac{1\text{ L}}{{10}^3\text{ mL}}\right)\\ =7.9341\times {10}^{-5}\text{ mol}\end{array}$

Substitute $150\text{ Torr}$ for $P$, $25.0\text{ mol}$ for $n$, $62.364\text{ L}\cdot \text{Torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$ and $298.15\text{K}$ for $T$ in equation (1) to calculate moles of $\text{HCl}$ gas.

  $\begin{array}{c}n=\left(\frac{\left(150\text{ Torr}\right)\left(25.0\text{ mL}\right)}{\left(62.364\text{ L}\cdot \text{Torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(298.15\text{K}\right)}\right)\left(\frac{1\text{ L}}{{10}^3\text{ mL}}\right)\\ =2.0116\times {10}^{-4}\text{ mol}\end{array}$

Since $\text{HCl}$ gas present in excess in reaction mixture thus ${\text{NH}}_3$ is a limiting reagent. Therefore, product is formed according to limiting reagent.

According to reaction, 1 mole of ${\text{NH}}_3$ produced by 1 mole of ${\text{NH}}_4\text{Cl}$ thus $7.9341\times {10}^{-5}\text{ mol}$ of ${\text{NH}}_4\text{Cl}$ can be produced from $7.9\times {10}^{-5}\text{ mol}$ of ${\text{NH}}_3$.

Mass of $7.9341\times {10}^{-5}\text{ mol}$ of ${\text{NH}}_4\text{Cl}$ can be calculated as follows:

  $\begin{array}{c}\text{Mass}\left({\text{NH}}_4\text{Cl}\right)=\left(\text{Moles}\left({\text{NH}}_4\text{Cl}\right)\right)\left(\text{Molar mass}\left({\text{NH}}_4\text{Cl}\right)\right)\\ =\left(7.9341\times {10}^{-5}\text{ mol}\right)\left(53.49\text{ g/mol}\right)\\ =4.25\times {10}^{-3}\text{ g}\end{array}$

Hence mass of product ${\text{NH}}_4\text{Cl}$ produced in reaction is $4.25\times {10}^{-3}\text{ g}$.

(b)

Interpretation Introduction

Interpretation:

Pressure of excess gas at $\text{27 }°\text{C}$ after completion of reaction has to be determined.

Concept Introduction:

Ideal gas law can be represented as equation for state for any imaginary gas. Expression for ideal gas equation is as follows:

  $PV=nRT$

Here,

$P$ is pressure of the gas.

$V$ is volume of gas.

$n$ denotes moles of a gas.

$R$ is gas constant.

$T$ is temperature of gas.

Answer to Problem 3C.15E

Pressure of excess gas at $\text{27 }°\text{C}$ after completion of reaction is $0.075\text{atm}$.

Explanation of Solution

The number of moles of excess $\text{HCl}\left(\text{g}\right)$ after reaction is completed can be calculated as follows:

  $\begin{array}{c}\text{Mole}\left(\text{HCl}\right)=\left(2.0116\times {10}^{-4}\text{ mol}\right)-\left(7.9341\times {10}^{-5}\text{ mol}\right)\\ =1.2181\times {10}^{-4}\text{ mol}\end{array}$

Total volume of mixture after the completion of reaction can be calculated as follows:

  $\begin{array}{c}\text{Total Volume of reaction mixture}=\left(25.0\text{ mL}\right)+\left(15.0\text{ mL}\right)\\ =40.0\text{ mL}\end{array}$

The expression to calculate pressure of gas is as follows:

  $P=\frac{nRT}{V}$        (2)

Conversion factor to convert $27\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273.15\\ =27\text{°C}+273.15\\ =300.15\text{K}\end{array}$

Substitute $40.0\text{ mL}$ for $V$, $1.2181\times {10}^{-4}\text{ mol}$ for $n$, $62.364\text{ L}\cdot \text{Torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$ and $300.15\text{K}$ for $T$ in equation (1) to calculate volume of $\text{HCl}$ gas.

  $\begin{array}{c}P=\left(\frac{\left(1.2181\times {10}^{-4}\text{ mol}\right)\left(62.364\text{ L}\cdot \text{Torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(300.15\text{K}\right)}{\left(40.0\text{ mL}\right)}\right)\left(\frac{{10}^3\text{ mL}}{1\text{ L}}\right)\\ =57.0068\text{ Torr}\left(\frac{\text{1}\text{atm}}{760\text{Torr}}\right)\\ =0.075\text{atm}\end{array}$

Hence pressure of excess gas at $\text{27 }°\text{C}$ after completion of reaction is $0.075\text{atm}$.