A halothane-oxygen mixture (C2 HBrClF3+O2) can be used as an anesthetic. A tank containing such

step 1 In this question, we have a halothane oxygen mixture. I'm going to abbreviate halothan as hal, o

A halothane-oxygen mixture (C2 HBrClF3+O2) can be used as an...

A halothane-oxygen mixture $\left(\mathrm{C}_{2} \mathrm{HBrClF}_{3}+\mathrm{O}_{2}\right)$ can be used as an anesthetic. A tank containing such a mixture has the following partial pressures: $P$ (halothane) $=$ $170 \mathrm{mm} \mathrm{Hg}$ and $P\left(\mathrm{O}_{2}\right)=570 \mathrm{mm} \mathrm{Hg}$ (a) What is the ratio of the number of moles of halothane to the number of moles of $\mathrm{O}_{2} ?$ (b) If the tank contains $160 \mathrm{g}$ of $\mathrm{O}_{2},$ what mass of $\mathrm{C}_{2} \mathrm{HBrClF}_{3}$ is present?

A halothane-oxygen mixture $\left(\mathrm{C}_{2} \mathrm{HBrClF}_{3}+\mathrm{O}_{2}\right)$ can be used as an anesthetic. A tank containing such a mixture has the following partial pressures: $P$ (halothane) $=$
$170 \mathrm{mm} \mathrm{Hg}$ and $P\left(\mathrm{O}_{2}\right)=570 \mathrm{mm} \mathrm{Hg}$
(a) What is the ratio of the number of moles of halothane to the number of moles of $\mathrm{O}_{2} ?$
(b) If the tank contains $160 \mathrm{g}$ of $\mathrm{O}_{2},$ what mass of $\mathrm{C}_{2} \mathrm{HBrClF}_{3}$ is present?

Key Concepts

Dalton's Law of Partial Pressures

This law states that in a mixture of non-reacting gases, the total pressure is the sum of the partial pressures of individual gases. Each gas exerts a pressure independent of the others, and the partial pressure of a gas in the mixture is directly proportional to its mole fraction when temperature and volume are constant.

Ideal Gas Law and Mole Ratios

Under constant temperature and volume, the number of moles of a gas is directly proportional to its pressure. This relationship allows us to equate ratios of partial pressures to ratios of moles. It is a fundamental principle for determining the quantitative composition of gas mixtures.

Mass-to-Mole Conversions

Converting between mass and moles is a key concept in stoichiometry. It involves using the molar mass (the mass of one mole of a substance) to convert a given mass of a substance into the number of moles, which is essential for comparing quantities in chemical reactions or mixtures.

Transcript

00:01 In this question, we have a halothane oxygen mixture.

00:07 I'm going to abbreviate halothan as hal, okay, h -a -l, plus oxygen.

00:15 In this mixture, the partial pressure of halogen, sorry, of halothane, is 170 millimeters of mercury, and the partial pressure of oxygen, is 570 millimeters of mercury.

00:45 We need to calculate the ratio of the number of moles of halothane to the number of moles of oxygen.

00:53 In other words, we just need to calculate how many moles of halogen there are in the mix relative to the number of moles of oxygen.

01:08 Okay? now, one way to to do this is a very long way, using the equation of the ideal gas, we can find the number of malls of halothane by using the pressure, the partial pressure of halothane, the volume in the container, the volume of the container in which the mix is contained, the constant of ideal gas and the temperature in kelvin's.

01:48 However, we don't have the volume nor the temperature.

01:57 Therefore, how do we do this? we need to calculate the number of moles of halogen, which is the pressure of halogen, times the volume of the container, which is the same for both, for the mix, and for both gases, divided by the constant of the ideal gas, times the temperature, which is the same for both.

02:28 Therefore, this is what i'm going to put here.

02:34 We need the pressure of halothane times the volume of the container, divided by rt.

02:45 For oxygen, we need exactly the same thing.

02:49 We need n of oxygen could be the partial pressure of oxygen, times the volume divided by rt, which is so the volume is the same, r is the same, t is the same.

03:04 Partial pressure of oxygen times volume divided by rt.

03:09 Therefore, if all of this is the same, this would be cancelled out, and the mole ratio will be halogen to oxygen, to the mold of oxygen, would be 170, divided by 570.

03:35 170 divided by 570.

03:40 And the answer is 0 .3.

03:50 So we have 0 .3 moles of halothane per every mole of oxygen.

04:01 Okay? we can also say that we have, if i want to to have a whole number on the top.

04:21 And we have to, no, it's better to leave it like this.

04:28 This is the mold ratio.

04:30 Ok.

04:33 This answers a, number a.

04:37 Now we need to answer b.

04:39 In b, we need to calculate the mass of halothane if we know that the mass of oxygen is 160 grams...