Question

What volume of `"dihydrogen gas"` under a temperature of `723*K`, and a pressure of `745.3*mm*Hg`, is required to reduce a `34.21*g` mass of `"ferric oxide"` to iron metal?

Answer 1

`Fe_2O_3(s) + 3H_2(g) +Deltararr 2Fe(s) + 3H_2O(g)`

I make it a volume of `13*L` of dihydrogen gas.....

Explanation:

Well you got a stoichiometrically balanced equation where garbage OUT is necessarily equal to garbage in.

We gots a molar quantity of `(34.21*g)/(159.69*g*mol^-1)=0.214*mol` with respect to `"ferric oxide"`.....

And so we need `3xx0.214*mol=0.643*mol` with respect to dihydrogen gas....

And so we solve the Ideal Gas equation knowing that `1*atm-=760*mm*Hg` (i.e. that one atmosphere will support a column of mercury that is `760*mm` high.....)

We solve for.......

`V=(nRT)/P=(0.214*molxx0.0821*(L*atm)/(K*mol)xx723*K)/((745.3*mm*Hg)/(760*mm*Hg*atm^-1))`

`=???*L`

What is the mass of the dihydrogen gas....