Chapter 9, Problem 148QP

(a)

Interpretation Introduction

Interpretation:

The moles and mass of neon gas contained in a $20.0\text{ L}$ container at $389\text{ torr}$ and $300\text{ K}$ are to be calculated.

Explanation of Solution

The number of moles of a gas $\left(n\right)$ is equal to the ratio of the amount of gas $\left(m\right)$ present in the container to the molar mass $\left(MM\right)$ of that gas.

$n=\frac{m}{MM}$ … (1)

The ideal gas law dictates the behavior of ideal gases. It states that the pressure $\left(P\right)$ and volume $\left(V\right)$ of a gas is directly proportional to the number of moles $\left(n\right)$ present in the gas and the temperature $\left(T\right)$ of the gas.

$PV=nRT$ … (2)

Here, $R$ is the universal gas constant.

Convert $389\text{ torr}$ to $\text{atm}$ as shown below:

$\begin{array}{c}\text{760 torr}=1\text{ atm}\\ 389\text{ torr}=\frac{1\text{ atm}}{760\text{ torr}}\times 389\text{ torr}\\ =0.512\text{ atm}\end{array}$

Substitute $P$ as $0.512\text{ atm}$ , $T$ as $300\text{ K}$ , $R$ as $0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}$ and $V$ as $20.0\text{ L}$ in equation (2) to determine $n$ of neon gas:

$\begin{array}{c}0.512\text{ atm}\times 20\text{ L}=n\times 0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}\times 300.0\text{ K}\\ n=\frac{0.512\text{ atm}\times 20\text{ L}}{0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}\times 300.0\text{ K}}\\ =0.416\text{ moles}\end{array}$

So, $0.416\text{ mol}$ of neon gas are present. The mass of neon gas is determined by using equation (1). Substitute $n$ as $0.416\text{ mol}$ and $MM$ as $20.18{\text{ g mol}}^{-1}$ in equation (1):

$\begin{array}{c}0.416\text{ mol}=\frac{m}{20.18{\text{ g mol}}^{-1}}\\ m=0.416\text{ mol}\times 20.18{\text{ g mol}}^{-1}\\ =8.39\text{ g}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The moles and mass of hydrogen gas contained in a $20.0\text{ L}$ container at $389\text{ torr}$ and $300\text{ K}$ are to be calculated.

Explanation of Solution

Substitute $P$ as $0.512\text{ atm}$ , $T$ as $300\text{ K}$ , $R$ as $0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}$ and $V$ as $20.0\text{ L}$ in equation (2) to determine $n$ of hydrogen gas:

$\begin{array}{c}0.512\text{ atm}\times 20\text{ L}=n\times 0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}\times 300.0\text{ K}\\ n=\frac{0.512\text{ atm}\times 20\text{ L}}{0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}\times 300.0\text{ K}}\\ =0.416\text{ moles}\end{array}$

So, $0.416\text{ mol}$ of hydrogen gas are present. The mass of hydrogen gas is determined by using equation (1). The molecular mass of hydrogen gas is:

$\begin{array}{c}{\left(MM\right)}_{{\text{H}}_{\text{2}}}=2\times 1{\text{ g mol}}^{-1}\\ =2{\text{ g mol}}^{-1}\end{array}$

Substitute $n$ as $0.416\text{ mol}$ and $MM$ as $2{\text{ g mol}}^{-1}$ in equation (1) :

$\begin{array}{c}0.416\text{ mol}=\frac{m}{2{\text{ g mol}}^{-1}}\\ m=0.416\text{ mol}\times 2{\text{ g mol}}^{-1}\\ =0.839\text{ g}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The moles and mass of carbon dioxide gas contained in a $20.0\text{ L}$ container at $389\text{ torr}$ and $300\text{ K}$ are to be calculated.

Explanation of Solution

Substitute $P$ as $0.512\text{ atm}$ , $T$ as $300\text{ K}$ , $R$ as $0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}$ and $V$ as $20.0\text{ L}$ in equation (2) to determine $n$ of carbon dioxide gas:

$\begin{array}{c}0.512\text{ atm}\times 20\text{ L}=n\times 0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}\times 300.0\text{ K}\\ n=\frac{0.512\text{ atm}\times 20\text{ L}}{0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}\times 300.0\text{ K}}\\ =0.416\text{ moles}\end{array}$

So, $0.416\text{ mol}$ of carbon dioxide gas are present. The mass of carbon dioxide gas is determined by using equation (1). The molecular mass of carbon dioxide gas is:

$\begin{array}{c}{\text{CO}}_2=12{\text{ g mol}}^{-1}+2\left(16{\text{ g mol}}^{-1}\right)\\ =44{\text{ g mol}}^{-1}\end{array}$

Substitute $n$ as $0.416\text{ mol}$ , $MM$ as ${\text{44 g mol}}^{-1}$ in equation (1) :

$\begin{array}{c}0.416\text{ mol}=\frac{m}{{\text{44 g mol}}^{-1}}\\ m=0.416\text{ mol}\times 44{\text{ g mol}}^{-1}\\ =18.3\text{ g}\end{array}$