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

Jun 18, 2018

We ASSUME that the aluminum reacts with EXCESS $H C l \left(a q\right)$...and get a volume of under $3 \cdot L$...

#### Explanation:

And so we address the redox reaction....

$A l \left(s\right) + 3 H C l \left(a q\right) \rightarrow A l C {l}_{3} \left(a q\right) + \frac{3}{2} {H}_{2} \left(g\right) \uparrow$

$\text{Moles of metal} \equiv \frac{2.00 \cdot g}{27.0 \cdot g \cdot m o {l}^{-} 1} = 0.0741 \cdot m o l$...

And so the reaction should produce $\frac{3}{2} \cdot \text{equiv}$ dihydrogen gas, i.e. $0.111 \cdot m o l$....and this represents a VOLUME under standard conditions*...of...

$0.111 \cdot m o l \times {\underbrace{24.5 \cdot L \cdot m o {l}^{-} 1}}_{\text{molar volume at 298 K and 1 atm}} = 2.72 \cdot L$

(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!

Jun 18, 2018

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

$2 A l \left(s\right) + 6 H C l \left(a q\right) \to 2 A l C {l}_{3} \left(a q\right) + 3 {H}_{2} \left(g\right)$

Assuming that our acid is in excess,

$n = 2 \text{g" * "mol"/(27.0"g") * (3H_2)/(2Al) approx 0.111"mol}$

Now, recall,

$P V = n R T$

Hence,

$\implies V = \frac{n R T}{P} \approx 2.50 \text{L}$

of hydrogen gas will be evolved from the reaction vessel.