Gaseous iodine pentafluoride, \(\mathrm{IF}_{5}\), can be prepared by the reaction of solid iodine and gaseous fluorine: $$\mathrm{I}_{2}(s)+5 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{IF}_{5}(g)$$ 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?

Step by step solution

01

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: \(n_{\mathrm{I}_2} = \frac{10.0 \mathrm{~g}}{253.8 \frac{\mathrm{g}}{\mathrm{mol}}} = 0.0394\, \mathrm{mol}\) \(n_{\mathrm{F}_2} = \frac{10.0 \mathrm{~g}}{38 \frac{\mathrm{g}}{\mathrm{mol}}} = 0.263\, \mathrm{mol}\)

02

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: \(0.0394 \,\mathrm{mol}\, \mathrm{I}_2 × \frac{5\, \mathrm{mol}\,\mathrm{F}_2}{1\,\mathrm{mol}\,\mathrm{I}_2} = 0.197\, \mathrm{mol}\,\mathrm{F}_2\) Since we only have \(0.263\, \mathrm{mol}\, \mathrm{F}_2\), we have sufficient \(\mathrm{F}_2\). So, the limiting reagent is \(\mathrm{I}_2\).

03

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: \(0.0394\, \mathrm{mol}\, \mathrm{I}_2 × \frac{2\, \mathrm{mol}\, \mathrm{IF}_5}{1\, \mathrm{mol}\, \mathrm{I}_2} = 0.0788\, \mathrm{mol}\, \mathrm{IF}_5\) Moles of \(\mathrm{F}_2\) remaining: \(0.263\, \mathrm{mol}\, \mathrm{F}_2 - 0.197\, \mathrm{mol}\, \mathrm{F}_2 = 0.066\, \mathrm{mol}\, \mathrm{F}_2\)

04

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 \(\mathrm{IF}_5\): \(P_{\mathrm{IF}_5} = \frac{n_{\mathrm{IF}_5}RT}{V} = \frac{0.0788\, \mathrm{mol} \times 0.0821\, \frac{\mathrm{L}\cdot \mathrm{atm}}{\mathrm{mol}\cdot \mathrm{K}} \times 398.15\, \mathrm{K}}{5.00\, \mathrm{L}} = 5.15 \, \mathrm{atm}\)

05

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 \(0.544\).

06

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

07

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 \(20.0\, \mathrm{g}\).

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