Question

Gaseous iodine pentafluoride, ${\mathrm{IF}}_5$, can be prepared by the reaction of solid iodine and gaseous fluorine: $${\mathrm{I}}_2(s)+5{\mathrm{F}}_2(g)⟶2{\mathrm{IF}}_5(g)$$ A $5.00-\mathrm{L}$ flask containing $10.0\mathrm{g}$ of ${\mathrm{I}}_2$ is charged with $10.0\mathrm{g}$ of ${\mathrm{F}}_2,$ and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is ${125}^{\circ }\mathrm{C}$. (a) What is the partial pressure of ${\mathrm{IF}}_5$ in the flask? (b) What is the mole fraction of ${\mathrm{IF}}_5$ in the flask (c) Draw the Lewis structure of IF ${}_5$. (d) What is the total mass of reactants and products in the flask?

Short answer

(a) The partial pressure of ${\mathrm{IF}}_5$ in the flask is $5.15\mathrm{atm}$. (b) The mole fraction of ${\mathrm{IF}}_5$ in the flask is $0.544$. (c) The Lewis structure of ${\mathrm{IF}}_5$ is: F - I - F | | F F | F (d) The total mass of reactants and products in the flask is $20.0\mathrm{g}$.

Step-by-step solution

Calculate the moles of reactants

First, calculate the moles of ${\mathrm{I}}_2$ and ${\mathrm{F}}_2$ in the reaction using their molar masses: Molar masses: ${\mathrm{I}}_2=253.8\frac{g}{mol}$ ${\mathrm{F}}_2=38\frac{g}{mol}$ Number of moles: $n_{{\mathrm{I}}_2}=\frac{10.0\mathrm{g}}{253.8\frac{\mathrm{g}}{\mathrm{mol}}}=0.0394\mathrm{mol}$ $n_{{\mathrm{F}}_2}=\frac{10.0\mathrm{g}}{38\frac{\mathrm{g}}{\mathrm{mol}}}=0.263\mathrm{mol}$

Identify the limiting reagent

To identify the limiting reagent, compare the mole ratio of the reactants: Mole ratio: $\frac{1}{5}$ (based on the balanced equation) Using the mole ratio, calculate the needed moles of ${\mathrm{F}}_2$ for all the iodine to react completely: $0.0394\mathrm{mol}{\mathrm{I}}_2\times \frac{5\mathrm{mol}{\mathrm{F}}_2}{1\mathrm{mol}{\mathrm{I}}_2}=0.197\mathrm{mol}{\mathrm{F}}_2$ Since we only have $0.263\mathrm{mol}{\mathrm{F}}_2$, we have sufficient ${\mathrm{F}}_2$. So, the limiting reagent is ${\mathrm{I}}_2$.

Find the moles of products and remaining reagent

Based on the limiting reagent (${\mathrm{I}}_2$), calculate the moles of ${\mathrm{IF}}_5$ produced and the moles of ${\mathrm{F}}_2$ remaining: Moles of ${\mathrm{IF}}_5$ produced: $0.0394\mathrm{mol}{\mathrm{I}}_2\times \frac{2\mathrm{mol}{\mathrm{IF}}_5}{1\mathrm{mol}{\mathrm{I}}_2}=0.0788\mathrm{mol}{\mathrm{IF}}_5$ Moles of ${\mathrm{F}}_2$ remaining: $0.263\mathrm{mol}{\mathrm{F}}_2-0.197\mathrm{mol}{\mathrm{F}}_2=0.066\mathrm{mol}{\mathrm{F}}_2$

Calculate the partial pressure of ${\mathrm{IF}}_5$

Graph the partial pressure of the gas using the ideal gas law equation: $PV=nRT$ Where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant ($0.0821\frac{\mathrm{L}\cdot \mathrm{atm}}{\mathrm{mol}\cdot \mathrm{K}}$), and $T$ is the temperature in Kelvin. To find the temperature in Kelvin: $T_{\mathrm{K}}={125}^{\circ }\mathrm{C}+273.15=398.15\mathrm{K}$ Now, find the partial pressure of ${\mathrm{IF}}_5$: $P_{{\mathrm{IF}}_5}=\frac{n_{{\mathrm{IF}}_5}RT}{V}=\frac{0.0788\mathrm{mol}\times 0.0821\frac{\mathrm{L}\cdot \mathrm{atm}}{\mathrm{mol}\cdot \mathrm{K}}\times 398.15\mathrm{K}}{5.00\mathrm{L}}=5.15\mathrm{atm}$

Calculate the mole fraction of ${\mathrm{IF}}_5$

Calculate the mole fraction of ${\mathrm{IF}}_5$: Mole fraction = $\frac{n_{{\mathrm{IF}}_5}}{n_{{\mathrm{IF}}_5}+n_{{\mathrm{F}}_2}}=\frac{0.0788\mathrm{mol}}{0.0788\mathrm{mol}+0.066\mathrm{mol}}=0.544$ So the mole fraction of ${\mathrm{IF}}_5$ is $0.544$.

Draw the Lewis structure of ${\mathrm{IF}}_5$

The Lewis structure of ${\mathrm{IF}}_5$ consists of one Iodine (I) atom, surrounded by five Fluorine (F) atoms. The Iodine atom shares one single bond with each of the Fluorine atoms. The Iodine atom has one lone pair of electrons. F - I - F | | F F | F

Calculate the total mass of reactants and products

As the amount of reactants has not changed and no extra substances have entered the flask, the total mass of reactants and products in the flask remains the same as the initial mass: Total mass $=10.0\mathrm{g}{\mathrm{I}}_2+10.0\mathrm{g}{\mathrm{F}}_2=20.0\mathrm{g}$ So, the total mass of reactants and products in the flask is $20.0\mathrm{g}$.

Key concepts

Limiting Reagent

The concept of the limiting reagent is crucial in stoichiometry. It determines how much product can be formed in a chemical reaction. In every chemical equation, reactants are consumed in a specific ratio, as dictated by the balanced equation. However, in real situations, these ratios aren't always perfectly met.
To find the limiting reagent, you need to compare the amount of reactants you have with the amount needed to completely react with other reactants, according to the stoichiometric coefficients in the balanced chemical equation.
In the iodine and fluorine reaction to produce IF extsubscript{5}, I extsubscript{2} is the limiting reagent. This is because the reaction requires a larger amount of fluorine relative to iodine (a 5:1 ratio), and we only have just enough fluorine to react with all the iodine present. Thus, iodine is consumed first, determining the maximum amount of IF extsubscript{5} that can be produced. Identifying the limiting reagent ensures we can calculate the amount of products accurately.

Ideal Gas Law

The ideal gas law is a key principle in chemistry that helps us find unknown characteristics about gases. This law is expressed as: $$PV=nRT$$Where:

• $P$ is the pressure of the gas,
• $V$ is the volume of the gas,
• $n$ is the number of moles of the gas,
• $R$ is the ideal gas constant, and
• $T$ is the temperature in Kelvin.

In the reaction producing IF extsubscript{5}, we used the ideal gas law to calculate the partial pressure of IF extsubscript{5} after the reaction. By knowing the moles of IF extsubscript{5} formed, the temperature, and the volume, we could rearrange the formula to solve for pressure.Given the conditions in the flask, this allows us to predict how gases behave under various conditions and ensures that calculations align closely with real-world measurements.

Mole Fraction

Mole fraction is a way of expressing the concentration of a component in a mixture, specifically a gaseous mixture in this context. It is defined as the ratio of the moles of a particular component to the total moles of all components in the mixture. It is denoted by the formula:Mole Fraction $X=\frac{n_i}{n_{\text{total}}}$Where:

• $n_i$ is the number of moles of the component of interest,
• $n_{\text{total}}$ is the total number of moles of all components in the mixture.

In the iodine pentafluoride production, the mole fraction helped us determine how much of the gas mixture was IF extsubscript{5} after the reaction concluded. Knowing this proportion is valuable for further chemical processes or reactions and can help in calculating properties like partial pressure based on the total pressure of the mixture.

Lewis Structure

Drawing a Lewis structure is a core skill in understanding the bonding and shape of molecules. It is a diagram that represents the valence electrons in a molecule. It helps visualize how atoms are connected and where the electrons are distributed. For IF extsubscript{5}, the Lewis structure represents iodine bonded to five fluorine atoms with single bonds. Iodine, which is in Group 17 of the periodic table, typically garners 7 valence electrons. In IF extsubscript{5}, iodine shares electrons with five fluorine atoms, each providing one electron for the bond, fulfilling iodine's expanded octet requirement (common for period 5 and beyond). Additionally, the iodine atom in IF extsubscript{5} retains one lone pair, which influences the molecular geometry, giving it a square pyramidal shape. This understanding helps in predicting the molecule's behavior and interactions in chemical reactions.