Question #c8a2e
1 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
Rearrange to solve for
#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,
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
#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.