Question

What volume of `H_2(g)` is produced when 2.00 g of `Al(s)` reacts at STP?

Answer 1

We ASSUME that the aluminum reacts with EXCESS `HCl(aq)`...and get a volume of under `3*L`...

Explanation:

And so we address the redox reaction....

`Al(s) + 3HCl(aq) rarr AlCl_3(aq) +3/2H_2(g)uarr`

`"Moles of metal"-=(2.00*g)/(27.0*g*mol^-1)=0.0741*mol`...

And so the reaction should produce `3/2*"equiv"` dihydrogen gas, i.e. `0.111*mol`....and this represents a VOLUME under standard conditions*...of...

`0.111*molxxunderbrace(24.5*L*mol^-1)_"molar volume at 298 K and 1 atm"=2.72*L`

(this is actually a little bit too much to collect in an upturned graduated cylinder!).

*what are standard conditions? You got me...they are all over the place these days....I know of THREE definitions in THREE national curricula...for God's sake!

Answer 2

You're probably referring to the reduction-oxidation reaction of aluminum metal in aqueous acid,

`2Al(s) + 6HCl(aq) to 2AlCl_3(aq) + 3H_2(g)`

Assuming that our acid is in excess,

`n = 2"g" * "mol"/(27.0"g") * (3H_2)/(2Al) approx 0.111"mol"`

Now, recall,

`PV = nRT`

Hence,

`=> V = (nRT)/P approx 2.50"L"`

of hydrogen gas will be evolved from the reaction vessel.