# Question #c8a2e

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that you need to use the **ideal gas law** equation to determine how many moles of gas are present in that sample.

Once you know the **total number** of moles present in the sample, use the **molar masses** of the two gases to find a relationship between the *mass* of each gas and the **total mass** of the gaseous mixture.

So, the ideal gas law equation looks like this

#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "# , where

*universal gas constant*, usually given as

**absolute temperature** of the gas

Rearrange to solve for *total number of moles* of gas present in the mixture

#PV = nRT implies n = (PV)/(RT)#

Now, **do not** forget that the units **must match** those used in the expression of the ideal gas constant, *mmHg* to *atm* and the temperature from *degrees Celsius* to *Kelvin*

Use the conversion factors

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 atm " = " 760 mmHg")color(white)(a/a)|)))# #color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))#

Plug in your values to get

#n = (870.2/760color(red)(cancel(color(black)("atm"))) * 15.1color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 31.2)color(red)(cancel(color(black)("K"))))#

#n = "0.6919 moles"#

So, you know that the mixture contains a total of

Now, if you take **mass** each gas contributes to the total

#m_(N_2) = x color(red)(cancel(color(black)("moles N"_2))) * "28.01 g"/(1color(red)(cancel(color(black)("mole N"_2)))) = 28.01 * x color(white)(a)"g"#

#m_(CO_2) = y color(red)(cancel(color(black)("moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = 44.01 * ycolor(white)(a)"g"#

You can now write two equations with two unknowns

#x + y = 0.6919#

#28.01 * x + 44.01 * y = 24.1#

This will get you

#x = 0.6919-y#

#28.01 * (0.6919 - y) + 44.01y = 24.1#

#19.38 - 28.01 y + 44.01y = 24.1#

#16y = 4.72 implies y= 4.72/16 = 0.295#

The value of

#x = 0.6919 - 0.295 = 0.3969#

So, the mixture contains

To get the **mole fraction** of nitrogen gas, divide the number of moles of nitrogen gas by the **total number of moles8* present in the mixture

#chi_(N_2) = (0.3969 color(red)(cancel(color(black)("moles"))))/(0.6919color(red)(cancel(color(black)("moles")))) = color(green)(|bar(ul(color(white)(a/a)0.574color(white)(a/a)|)))#

The answer is rounded to three **sig figs**.