Question

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`?

Answer 1

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`