Question #cc3ae
1 Answer
Explanation:
The first thing that you need to do here is to figure out the mole ratio that exists between the two gases in the mixture.
To do that, use the molar masses of hydrogen gas and of oxygen gas.
M_ ("M O"_ 2) = "32.0 g mol"^(-1)MM O2=32.0 g mol−1
M_ ("M H"_ 2) ~~ "2.0 g mol"^(-1)MM H2≈2.0 g mol−1
Now, you know that this mixture contains
m_ ("H"_ 2) = 20/100 * mcolor(white)(.)"g" = (1/5 * m)color(white)(.)"g"mH2=20100⋅m.g=(15⋅m).g
m_ ("O"_ 2) = mcolor(white)(.)"g" - (1/5 * m) color(white)(.)"m" = (4/5 * m)color(white)(.)"g"mO2=m.g−(15⋅m).m=(45⋅m).g
The number of moles of each gas present in the mixture will be
(1/5 * m) color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.0color(red)(cancel(color(black)("g")))) = (1/10 * m)color(white)(.)"moles H"_2
(4/5 * m) color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = (1/40 * m)color(white)(.)"moles O"_2
Now, you know that the partial pressure of a gas that's part of a gaseous mixture depends on its mole fraction in the mixture and on the total pressure of the mixture.
The mole fraction of hydrogen gas will be equal to the number of moles of hydrogen gas divided by the total number of moles present in the mixture.
chi_ ("H"_ 2) = ((1/10 * color(red)(cancel(color(black)(m)))) color(red)(cancel(color(black)("moles"))))/((1/10 * color(red)(cancel(color(black)(m)))+ 1/40 * color(red)(cancel(color(black)(m)))) color(red)(cancel(color(black)("moles")))) = 1/10 * 40/5 = 4/5
If you take
color(blue)(ul(color(black)(P_ ("H"_ 2) = chi_ ("H"_ 2) * P_"total")))
P_ ("H"_ 2) = 4/5 * P_"total"
In your case, you have
P_"total" = "1 bar"
which means that the partial pressure of hydrogen gas will be
P_ ("H"_ 2) = 4/5 * "1 bar" = color(darkgreen)(ul(color(black)("0.8 bar")))
The answer is rounded to one significant figure.
