# How is aluminum oxidized under basic conditions? What is the reduction product?

## What volume of this product is collected under standard conditions when a $5.4 \cdot g$ mass of this metal is oxidized in a solution prepared from a $20 \cdot g$ mass of sodium hydroxide?

Mar 16, 2017

Approx. $7 \cdot L$.

#### Explanation:

We interrogate the oxidation reaction:

$A l + 4 H {O}^{-} \rightarrow A l {\left(O H\right)}_{4}^{-} + 3 {e}^{-}$ $\left(i\right)$

And water is reduced to dihydrogen gas:

${H}_{2} O + {e}^{-} \rightarrow \frac{1}{2} {H}_{2} \left(g\right) + H {O}^{-}$ $\left(i i\right)$

And so, we take $\left(i\right) + 3 \times \left(i i\right) :$

$A l + H {O}^{-} + 3 {H}_{2} O \rightarrow A l {\left(O H\right)}_{4}^{-} + \frac{3}{2} {H}_{2} \left(g\right)$

We have the starting masses, and thus:

$\text{Moles of aluminum} = \frac{5.4 \cdot g}{26.98 \cdot g \cdot m o {l}^{-} 1} = 0.200 \cdot m o l$.

$\text{Moles of NaOH} = \frac{20.0 \cdot g}{40.00 \cdot g \cdot m o {l}^{-} 1} = 0.500 \cdot m o l$.

Clearly, hydroxide is the reagent in excess...........as required.

And given the stoichiometry, there are $0.300 \cdot m o l$ dihydrogen gas evolved. At $\text{STP}$, the molar volume is $22.4 \cdot L \cdot m o {l}^{-} 1$.

And thus the dihydrogen gas evolved has a volume of 0.300*molxx22.4*L*mol^-1=??L.

Aluminum is amphoteric and can be oxidized under acid, or under basic conditions (as here). I take it this is 1st year undergrad?