When #"4 g H"_2# and #"32 g O"_2# are mixed, what is the partial pressure of oxygen at a total pressure of #P#?

About #2/3# of #P#.

Without your calculator, what is the fraction of #P# that oxygen gas exerts?

For ideal gases, the partial pressure is given by

#P_i = chi_iP#,

where:

• #P# is the total pressure.
• #P_i# is the pressure of gas #i# by itself, assuming it is not interacting with anything else.
• #chi_i = (n_i)/(n_1 + n_2 + . . . + n_N)# is the mol fraction of gas #i# in the container.
• #n_i# is the mols of gas #i#.

The fraction of the total pressure #P# is given by

#chi_i = P_i/P#

And so, all we need to do is find the mols of each gas and calculate the mol fraction of #"H"_2#, i.e.:

#chi_(H_2) = (n_(H_2))/(n_(H_2) + n_(O_2))#

The mols of #"H"_2# are given by:

#4 cancel("g H"_2) xx "1 mol H"_2/(2.0158 cancel("g H"_2))#

#=# #"1.984 mols"#

The mols of #"O"_2# are given by:

#32 cancel("g O"_2) xx "1 mol O"_2/(31.998 cancel("g O"_2))#

#=# #"1.000 mols"#

Therefore, the fraction of #P# that #"H"_2# exerts is given by

#color(blue)(chi_(H_2)) = ("1.984 mols")/("1.984 mols H"_2 + "1.000 mols O"_2)#

#= color(blue)(0.665)#

This is about #color(blue)(2/3)# #color(blue)("of")# #color(blue)(P)#.