If #80*g# of magnesium chloride salt were isolated from the treatment of magnesium metal by excess hydrochloric acid, how much metal was present?

1 Answer
May 5, 2017

Answer:

We interrogate the stoichiometric equation..........and find that approx. #20*g# of magnesium was present.

Explanation:

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

If there are #80*g# salt produced, this represents a molar quantity of #(80*g)/(95.21*g*mol^-1)=0.840*mol#.

And thus there MUST have been an #0.840*mol# quantity of metal present, given the 1:1 stoichiometry of the reaction. And so we gots a mass with respect to the metal of...

#0.840*gxx24.31*g*mol^-1=??*g#

And thus there were also a #0.840*mol# quantity of dihydrogen gas evolved. If the gas is (reasonably!) assumed to behave ideally, then a volume of #0.840*molxx24.5*L*mol^-1=20.6*L# under standard condtions of #1*atm#, and #298*K# is evolved.