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"Moles of dihydrogen == (3.7*g)/(2.02*g*mol^-1)=1.83*mol3.7⋅g2.02⋅g⋅mol−1=1.83⋅mol.
"Moles of dioxygen"Moles of dioxygen == (9.1*g)/(32.00*g*mol^-1)=4.97*mol9.1⋅g32.00⋅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*atmPdihydrogen=318⋅mm⋅Hg=318⋅mm⋅Hg760⋅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*atmPdioxygen=4.97⋅mol1.83⋅mol×0.418⋅atm=1.135⋅atm.
And finally, P_"Total"=P_"dihydrogen"+P_"dioxygen"PTotal=Pdihydrogen+Pdioxygen
== (0.418+1.135)*atm(0.418+1.135)⋅atm
Note that I converted the mm*Hgmm⋅Hg to units of "atmospheres"atmospheres, because I really don't think mm*Hgmm⋅Hg should be used for pressures much above 1*atm, i.e. 760*mm*Hg1⋅atm,i.e.760⋅mm⋅Hg. There seems to be quite a few questions on these boards where mercury columns greater than 760*mm*Hg760⋅mm⋅Hg have been used to measure pressure. Obviously, the person who set the question has never had to clean up a mercury spill.
