How many liters of #O_2# gas, measured at 777 mm Hg and 35°C, are required to completely react with 2.7 mol of #Al#?

1 Answer
anor277
Nov 19, 2016

Approx. #50*L# of #"dioxygen gas"#

Explanation:

A measurement of #777" mm Hg"# is completely unrealistic. Given #1*atm-=760*mm*Hg#, a column of mercury that is longer than #760*mm# risks getting mercury all over the shop, and this is an outcome that you should avoid.

Given #777" mm Hg"# #=# #(777*mm*Hg)/(760*mm*Hg*atm^-1)=1.02*atm,# we need a stoichiometric equation:

#2Al(s) + 3/2O_2(g) rarr Al_2O_3(g)#

So if there are #2.7*mol# of metal, we need #2.025*mol# of #"dioxygen gas"#.

#V=(nRT)/P# #=# #(2.025*molxx0.0821*L*atm*K^-1*mol^-1xx308*K)/(1.02*atm)# #=??*L#

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