# What would occur if a 20*kg block of magnesium metal were dropped in a 500*L vat of conc. hydrochloric acid? What volume of dihydrogen gas would be generated?

Apr 12, 2017

And who did it? What a klutz............this is a major accident.

#### Explanation:

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

Conc. $H C l \left(a q\right)$, the stuff you get in a lab is $\text{32-34%(m/m)}$, and we will treat this as a molar concentration of $10.6 \cdot m o l \cdot {L}^{-} 1$ with respect to $H C l$. (How do I know this? Well, you need the density of the acid to make the calculation, but I have extensively used conc. hydrochloric acid and happen to know).

And thus moles of $H C l$:

$10.6 \cdot m o l \cdot {L}^{-} 1 \times 500 \cdot L = 5300 \cdot m o l$

And moles of $M g$:

$\frac{20 \times {10}^{3} \cdot g}{24.3 \cdot g \cdot m o {l}^{-} 1} = 823 \cdot m o l$ of metal...........

Given the stoichiometric equation, we would generate $823 \cdot m o l$ of $\text{dihydrogen gas}$.

And this has a volume of $V = \frac{n R T}{P}$ assuming ideality.

$V = \frac{823 \cdot m o l \times 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 305 \cdot K}{1 \cdot a t m}$

$\cong 20 \times {10}^{3} \cdot L$, i.e over $20 \cdot {m}^{3}$. This is a HUGE VOLUME, and would constitute a significant safety hazard (both in terms of combustion and asphyxiation!). What do you think would happen if someone (probably the klutz who dropped the metal!) lit up a cigarette?