Chapter 5.4, Problem 5.13BFP

Interpretation Introduction

Interpretation:

The volume of gaseous iodine pentafluoride measured $105\text{°C}$ and $0.935\text{atm}$ prepared by the reaction of $4.16\text{g}$ of solid iodine with $2.48\text{L}$ of gaseous fluorine at $18°\text{C}$ and $0.974\text{atm}$ is to be calculated.

Concept introduction:

Ideal gas law equation is as follows:

$PV=nRT$ (1)

Here,

$P$ is the pressure of the gas.

$V$ is the volume occupied by gas.

$n$ is the number of moles of gas.

$T$ is the temperature in Kelvin.

$R$ is the universal gas constant.

Rearrange equation (1) for $n$ as follows:

$n=\frac{PV}{RT}$ (2)

Conversion factor to convert $\text{mL}$ to $\text{L}$ is as follows:

$1\text{mL}={10}^{-3}\text{L}$

The conversion factor to convert $°\text{C}$ to Kelvin is as follows:

$T\left(\text{K}\right)=T\left(\text{°C}\right)+273.15$ (3)

When the number of each atom in the reactant side is equal toa number of each atom in the product side then the reaction is known as a balanced chemical reaction. The reactant present in the lower quantity of moles is the limiting reagent.

Answer to Problem 5.13BFP

The volume of gaseous iodine pentafluoride measured $105\text{°C}$ and $0.935\text{atm}$ prepared by the reaction of $4.16\text{g}$ of solid iodine with $2.48\text{L}$ of gaseous fluorine at $18°\text{C}$ and $0.974\text{atm}$ is $\text{1}\text{.09}\text{L}$.

Explanation of Solution

The value of the universal gas constant $\left(R\right)$ is $\text{0}\text{.0821}\text{atm}\cdot \text{litre/mol}\cdot \text{K}$.

The pressure of fluorine gas $\left(P_{{\text{F}}_{\text{2}}}\right)$ is $0.974\text{atm}$.

 The volume $\left(V\right)$ of the flask containing fluorine gas is $\text{2}\text{.48}\text{L}$.

The weight of iodine gas is $4.16\text{g}$.

The molecular weight of ${\text{I}}_{\text{2}}$ is $253.8\text{g/mol}$.

The value of the temperature of fluorine gas is $18°\text{C}$.

The value of the temperature of ${\text{IF}}_{\text{5}}$ is $105°\text{C}$.

Substitute $18°\text{C}$ for in equation (3) to calculate the temperature in Kelvin.

$\begin{array}{c}T\left(\text{K}\right)=\left(18°\text{C}+273.15\right)\\ =291.15\text{K}\end{array}$

Substitute $105°\text{C}$ for in equation (3) to calculate the temperature in Kelvin.

$\begin{array}{c}T\left(\text{K}\right)=\left(105°\text{C}+273.15\right)\\ =378.15\text{K}\end{array}$

The balanced chemical equation for the given reaction is as follows:

${\text{I}}_{\text{2}}\left(s\right)+\text{5}{\text{F}}_{\text{2}}\left(g\right)\to {\text{2IF}}_{\text{5}}\left(g\right)$

Substitute $\text{0}\text{.0821}\text{atm}\cdot \text{litre/mol}\cdot \text{K}$ for $R$, $0.974\text{atm}$ for $P$, $\text{2}\text{.48}\text{L}$ for $V$, $291.15\text{K}$ for $T$ in equation (2) to calculate the number of moles of fluorine gas.

$\begin{array}{c}{n}_{{\text{F}}_{\text{2}}}=\frac{\left(0.974\text{atm}\right)\left(\text{2}\text{.48}\text{L}\right)}{\left(\frac{\text{0}\text{.0821}\text{atm}\cdot \text{litre}}{\text{mol}\cdot \text{K}}\right)\left(\text{291}\text{.15}\text{K}\right)}\\ =0.10105\text{mol}\end{array}$

According to the balanced chemical equation, the stoichiometric ratio between ${\text{IF}}_{\text{5}}$ and ${\text{F}}_{\text{2}}$ is 2:5.

The formula to calculate the number of moles of ${\text{IF}}_{\text{5}}$ from ${\text{F}}_{\text{2}}$ is as follows:

$n_{{\text{IF}}_{\text{5}}}=\left(n_{{\text{F}}_2}\right)\left(\frac{\text{2}\text{mol}{\text{IF}}_{\text{5}}}{\text{5}\text{mol}{\text{F}}_{\text{2}}}\right)$ (4)

Substitute $0.10105\text{mol}$ for $n_{{\text{F}}_{\text{2}}}$ in equation (4).

$\begin{array}{c}{n}_{{\text{IF}}_{\text{5}}}=\left(0.10105\text{mol}\right)\left(\frac{\text{2}\text{mol}{\text{IF}}_{\text{5}}}{\text{5}\text{mol}{\text{F}}_{\text{2}}}\right)\\ =0.040421\text{mol}\end{array}$

According to the balanced chemical equation, the stoichiometric ratio between ${\text{IF}}_{\text{5}}$ and ${\text{I}}_{\text{2}}$ is 2:1.

The formula to calculate the number of moles of ${\text{IF}}_{\text{5}}$ from ${\text{I}}_{\text{2}}$ is as follows:

$n_{{\text{IF}}_{\text{5}}}=\left(\frac{m_{{\text{I}}_{\text{2}}}}{M_{{\text{I}}_{\text{2}}}}\right)\left(\frac{\text{2}\text{mol}{\text{IF}}_{\text{5}}}{1\text{mol}{\text{I}}_{\text{2}}}\right)$ (5)

Substitute $4.16\text{g}$ for $m_{{\text{I}}_{\text{2}}}$ and $253.8\text{g/mol}$ for $M_{{\text{I}}_{\text{2}}}$ in equation (5).

$\begin{array}{c}{n}_{{\text{IF}}_{\text{5}}}=\left(\frac{4.16\text{g}}{253.8\text{g/mol}}\right)\left(\frac{\text{2}\text{mol}{\text{IF}}_{\text{5}}}{\text{1}\text{mol}{\text{I}}_{\text{2}}}\right)\\ =0.032782 \text{mol}\end{array}$

The amount of ${\text{IF}}_{\text{5}}$ produced from ${\text{I}}_{\text{2}}$ is less than the amount of ${\text{IF}}_{\text{5}}$ produced from ${\text{F}}_{\text{2}}$, therefore, ${\text{I}}_{\text{2}}$ acts as a limiting reagent and amount of ${\text{IF}}_{\text{5}}$ will be produced in accordance to the mole of ${\text{I}}_{\text{2}}$.

Rearrange equation (2) for the volume of ${\text{IF}}_{\text{5}}$ as follows:

$V_{{\text{IF}}_{\text{5}}}=\frac{nRT}{P}$ (6)

Substitute $0.032782 \text{mol}$ for $n$, $\text{0}\text{.0821}\text{atm}\cdot \text{litre/mol}\cdot \text{K}$ for  $R$, $378.15\text{K}$ for $T$ and $0.935\text{atm}$ for $P$ in equation (6).

$\begin{array}{c}{V}_{{\text{IF}}_{\text{5}}}=\frac{\left(0.0327mol\right)\left(\frac{\text{0}\text{.0821}\text{atm}\cdot \text{litre}}{\text{mol}\cdot \text{K}}\right)\left(378.15\text{K}\right)}{\left(0.935\text{atm}\right)}\\ =1.0885\text{L}\\ =\text{1}\text{.09}\text{L}\end{array}$

Conclusion

The volume of gaseous iodine pentafluoride measured $105\text{°C}$ and $0.935\text{atm}$ prepared by the reaction of $4.16\text{g}$ of solid iodine with $2.48\text{L}$ of gaseous fluorine at $18°\text{C}$ and $0.974\text{atm}$ is $\text{1}\text{.09}\text{L}$.