What is the total pressure (mmHg) of a gaseous mixture of 3.7 g of hydrogen gas and 9.1 g of oxygen gas in a 3.24-L flask if the partial pressure of hydrogen is 318 mmHg?

1 Answer
anor277
Nov 25, 2016

Approx. #1.5*atm#.

Explanation:

Given constant temperature, and volume, which we certainly have here (even though we don't know their magnitudes), pressure is proportional to the number of moles of gas, or vice versa.

#"Moles of dihydrogen"# #=# #(3.7*g)/(2.02*g*mol^-1)=1.83*mol#.

#"Moles of dioxygen"# #=# #(9.1*g)/(32.00*g*mol^-1)=4.97*mol#.

We are given that #P_"dihydrogen"=318*mm*Hg=(318*mm*Hg)/(760*mm*Hg*atm^-1)=0.418*atm#

And given the proportionality between moles and pressure,

#P_"dioxygen"=(4.97*mol)/(1.83*mol)xx0.418*atm=1.135*atm#.

And finally, #P_"Total"=P_"dihydrogen"+P_"dioxygen"#

#=# #(0.418+1.135)*atm#

Note that I converted the #mm*Hg# to units of #"atmospheres"#, because I really don't think #mm*Hg# should be used for pressures much above #1*atm, i.e. 760*mm*Hg#. There seems to be quite a few questions on these boards where mercury columns greater than #760*mm*Hg# have been used to measure pressure. Obviously, the person who set the question has never had to clean up a mercury spill.

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