Chapter 3, Problem 3C.13E

(a)

Interpretation Introduction

Interpretation:

Volume of product produced from $2.00{\text{ mol F}}_2$ in reaction given below has to be determined.

  ${\text{Cl}}_{\text{2}}\left(\text{g}\right)+{\text{F}}_{\text{2}}\left(\text{g}\right)\to 2\text{ClF}\left(\text{g}\right)$

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.13E

Volume of product produced from $2.00{\text{ mol F}}_2$ is $97.8\text{ L}$.

Explanation of Solution

Balance reaction is given as follows:

  ${\text{Cl}}_{\text{2}}\left(\text{g}\right)+{\text{F}}_{\text{2}}\left(\text{g}\right)\to 2\text{ClF}\left(\text{g}\right)$

According to reaction, 1 mole of ${\text{F}}_2$ gas produces 2 moles $\text{ClF}$ thus amount of $\text{ClF}$ produce form $2.00{\text{ mol F}}_2$ can be calculated as follows:

  $\begin{array}{c}\text{Mass}\left(\text{ClF}\right)=\text{Mole}\left({\text{F}}_2\right)\left(\frac{2\text{ mol ClF}}{1{\text{ mol F}}_2}\right)\\ =\left(2.00{\text{ mol F}}_2\right)\left(\frac{2\text{ mol ClF}}{1{\text{ mol F}}_2}\right)\\ =4.00{\text{ mol F}}_2\end{array}$

The expression to calculate volume of gas is as follows:

  $V=\frac{nRT}{P}$        (1)

Substitute $1.00\text{ atm}$ for $P$, $2.00\text{ mol}$ for $n$, $8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$ and $298\text{K}$ for $T$ in equation (1).

  $\begin{array}{c}V=\frac{\left(4.00\text{ mol}\right)\left(8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(298\text{K}\right)}{\left(1.00\text{ atm}\right)}\\ =97.8\text{ L}\end{array}$

Hence volume of product produced from $2.00{\text{ mol F}}_2$  is $97.8\text{ L}$.

(b)

Interpretation Introduction

Interpretation:

Volume of product produced from $2.00{\text{ mol F}}_2$ in reaction given below has to be determined.

  ${\text{Cl}}_{\text{2}}\left(\text{g}\right)+3{\text{F}}_{\text{2}}\left(\text{g}\right)\to 2{\text{ClF}}_3\left(\text{g}\right)$

Concept Introduction:

Refer to part (a).

Answer to Problem 3C.13E

Volume of product produced from $2.00{\text{ mol F}}_2$ is $32.5\text{ L}$.

Explanation of Solution

Balance reaction is given as follows:

  ${\text{Cl}}_{\text{2}}\left(\text{g}\right)+3{\text{F}}_{\text{2}}\left(\text{g}\right)\to 2{\text{ClF}}_3\left(\text{g}\right)$

According to reaction, 3 mole of ${\text{F}}_2$ gas produces 2 moles ${\text{ClF}}_3$ thus amount of ${\text{ClF}}_3$ produce form $2.00{\text{ mol F}}_2$ can be calculated as follows:

  $\begin{array}{c}\text{Mass}\left({\text{ClF}}_3\right)=\text{Mole}\left({\text{F}}_2\right)\left(\frac{2{\text{ mol ClF}}_3}{3{\text{ mol F}}_2}\right)\\ =\left(2.00{\text{ mol F}}_2\right)\left(\frac{2{\text{ mol ClF}}_3}{3{\text{ mol F}}_2}\right)\\ =1.33{\text{ mol F}}_2\end{array}$

The expression to calculate volume of gas is as follows:

  $V=\frac{nRT}{P}$        (1)

Substitute $1.00\text{ atm}$ for $P$, $1.33\text{ mol}$ for $n$, $8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$ and $298\text{K}$ for $T$ in equation (1).

  $\begin{array}{c}V=\frac{\left(1.33\text{ mol}\right)\left(8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(298\text{K}\right)}{\left(1.00\text{ atm}\right)}\\ =32.5\text{ L}\end{array}$

Hence volume of product produced from $2.00{\text{ mol F}}_2$ is $32.5\text{ L}$.

(c)

Interpretation Introduction

Interpretation:

Volume of product produced from $2.00{\text{ mol F}}_2$ in reaction given below has to be determined.

  ${\text{Cl}}_{\text{2}}\left(\text{g}\right)+5{\text{F}}_{\text{2}}\left(\text{g}\right)\to 2{\text{ClF}}_5\left(\text{g}\right)$

Concept Introduction:

Refer to part (a).

Answer to Problem 3C.13E

Volume of product produced from $2.00{\text{ mol F}}_2$ is $19.56\text{ L}$.

Explanation of Solution

Balance reaction is given as follows:

  ${\text{Cl}}_{\text{2}}\left(\text{g}\right)+5{\text{F}}_{\text{2}}\left(\text{g}\right)\to 2{\text{ClF}}_5\left(\text{g}\right)$

According to reaction, 5 mole of ${\text{F}}_2$ gas produces 2 moles ${\text{ClF}}_5$ thus amount of ${\text{ClF}}_5$ produce form $2.00{\text{ mol F}}_2$ can be calculated as follows:

  $\begin{array}{c}\text{Mass}\left({\text{ClF}}_5\right)=\text{Mole}\left({\text{F}}_2\right)\left(\frac{2{\text{ mol ClF}}_5}{5{\text{ mol F}}_2}\right)\\ =\left(2.00{\text{ mol F}}_2\right)\left(\frac{2{\text{ mol ClF}}_5}{5{\text{ mol F}}_2}\right)\\ =0.80{\text{ mol F}}_2\end{array}$

The expression to calculate volume of gas is as follows:

  $V=\frac{nRT}{P}$        (1)

Substitute $1.00\text{ atm}$ for $P$, $0.80\text{ mol}$ for $n$, $8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$ and $298\text{K}$ for $T$ in equation (1).

  $\begin{array}{c}V=\frac{\left(0.80\text{ mol}\right)\left(8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(298\text{K}\right)}{\left(1.00\text{ atm}\right)}\\ =19.56\text{ L}\end{array}$

Hence volume of product produced from $2.00{\text{ mol F}}_2$ is $19.56\text{ L}$.