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

Oct 27, 2016

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

Approx, $8 \cdot L$ of dihydrogen gas are evolved.

#### Explanation:

Aluminum metal will reduce the hydrogen in water under basic conditions to give the aluminate ion $A l {\left(O H\right)}_{4}^{-}$.

$\text{Moles of metal}$ $=$ $\frac{6.32 \cdot g}{26.98 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.234 \cdot m o l$.

Now given the stoichiometric equation, 3/2 equiv of dihydrogen gas evolve from each equiv metal, i.e. $0.351 \cdot m o l$.

Since at $\text{STP}$ $1$ $m o l$ of ideal gas occupies $22.4 \cdot L \cdot m o {l}^{-} 1$, we take the product, 0.351*molxx22.4*L*mol^-1=??L