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?

1 Answer
Apr 12, 2017

Answer:

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

Explanation:

We follow the stoichiometric equation........

#Mg(s) + 2HCl(aq) rarr MgCl_2(aq) + H_2(g)uarr#

Conc. #HCl(aq)#, the stuff you get in a lab is #"32-34%(m/m)"#, and we will treat this as a molar concentration of #10.6*mol*L^-1# with respect to #HCl#. (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 #HCl#:

#10.6*mol*L^-1xx500*L=5300*mol#

And moles of #Mg#:

#(20xx10^3*g)/(24.3*g*mol^-1)=823*mol# of metal...........

Given the stoichiometric equation, we would generate #823*mol# of #"dihydrogen gas"#.

And this has a volume of #V=(nRT)/P# assuming ideality.

#V=(823*molxx0.0821*L*atm*K^-1*mol^-1xx305*K)/(1*atm)#

#~=20xx10^3*L#, i.e over #20*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?