Gaseous iodine pentafluoride, IF5, can be prepared by the reaction of solid iodine and gaseous fluorine: I21s2 + 5 F21g2¡2 IF51g2 A 5.00-L flask containing 10.0 g of I2 is charged with 10.0 g of F2, and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is 125 °C. (a) What is the partial pressure of IF5 in the flask?

Video transcript

Hi everyone here we have a question asking us to consider the reaction bro. Mean liquid plus chlorine gasses forms bromine chloride gashes and we're told that a 2.0 liter flask containing 5.0 g of roaming is charged with 2.50 g of chlorine. After the reaction is complete, the temperature in the flask is 150 degrees Celsius, assuming that the reaction proceeds until one of the re agents is completely consumed. Calculate the partial pressure of bromine chloride in the flask. So first we need a balanced reaction. So we have browning liquid plus chlorine. Gashes forms roaming chloride gashes And if we look at our reactive side, we see that we have to browning and only one on our product side. So we're going to put a two in front of their. Now we have 10 chlorine on our product side. So we're going to put a five in front of our chlorine on a reacting side. To balance that out. So now we have 10 on both sides. So now we can calculate the molds of bromine chloride produced using both react ints. So we need the molar mass. So for browning we have to Times 79. equals 1 59.8 g per mole in chlorine. We have two tons, 35. Equal 70.9 g per mole. So now we're going to start with our 5. brands of browning. Times one mole of browning Divided by .8 g of grooming times two moles. A roaming chloride. And that is from our balanced reaction over one mole. A roaming and that equals 0.0 - six moles of browning chloride. Now if we look here are grams of brown meat are canceling out, our moles of bromine are canceling out, leaving us with moles of bromine chloride, which is exactly what we want. So now we're going to do the exact same thing with our chlorine. So we have 2.50 g of chlorine times one more of chlorine over 70.9 g of chlorine. Times two moles. A roaming chloride over five moles. The chlorine and our grams of chlorine are canceling out our molds of chlorine are canceling out and that equals 0. moles of roaming. Cool ride. And if we compare the two, we see that we get less when we use chlorine. So chlorine is our limiting reactant and it gives us our theoretical yield. So now we can calculate the pressure of roaming chloride using the ideal gas law. The ideal gas law gives us the formula PV equals NRT. Our volume equals 2.00 L. Our number of moles equals 0. moles. Our gas constant equal 0. to 06. Leaders times atmospheres over moles time kelvin and our temperature Equals 150. And that's in C. and we want to change it to Kelvin. So we're gonna add 273.15 and that equals .15 Kelvin. So now we're going to rearrange our PV equals NRT to solve for pressure. So we're going to divide both sides by B, which gives us P equals NRT over V. So P equals 0.0141 moles Times 0.08206 L times atmospheres over moles times kelvin times 423.15 Kelvin, divided by two leaders Equals 0.245 Atmospheres. And that is our final answer. Thank you for watching. Bye.