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

Question

Jeremiah Adams · Accepted Answer

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

#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.

Cooper Anderson · Answer

undefined To find the amount of magnesium metal present, you need to first determine the molar mass of magnesium chloride (MgCl2). Then, use stoichiometry to find the amount of magnesium metal that would produce 80g of magnesium chloride.