Question

What is the volume of `H_2` has formed at STP when 6.32g of `Al` reacts with excess `NaOH`?

Answer 1

`Al + NaOH + 3H_2O → Na^(+) + [Al(OH)_4]^(-) + 3/2 H_2 (g)uarr`

Approx, `8*L` of dihydrogen gas are evolved.

Explanation:

Aluminum metal will reduce the hydrogen in water under basic conditions to give the aluminate ion `Al(OH)_4^-`.

`"Moles of metal"` `=` `(6.32*g)/(26.98*g*mol^-1)` `=` `0.234*mol`.

Now given the stoichiometric equation, 3/2 equiv of dihydrogen gas evolve from each equiv metal, i.e. `0.351*mol`.

Since at `"STP"` `1` `mol` of ideal gas occupies `22.4*L*mol^-1`, we take the product, `0.351*molxx22.4*L*mol^-1=??L`