A #55*mL# volume of hydrochloric acid was poured on to an excess of magnesium turnings. What volume of dihydrogen gas evolved?

1 Answer
anor277
Jul 17, 2017

Well, one mole of gas at #"STP"# occupies #22.4*L#.......We do not know the concentration of hydrochloric acid.......

Explanation:

And thus we interrogate the redox reaction........

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

And thus the moles of magnesium is equivalent to the moles of dihydrogen gas if there is molar equivalence.

Now I happen to know that if #rho_"HCl"=1.18*g*mol^-1#, then the concentration of #HCl# is #10.17*mol*L^-1# (and such data SHOULD have been supplied with the question).

And so we have a molar quantity of #55xx10^-3*Lxx10.17*mol*L^-1=0.559*mol# with respect to #HCl#.

#"Moles of magnesium"=(0.865*g)/(24.305*g*mol^-1)=0.0356*mol#.

Magnesium metal is in excess, and we do not have a handle on the volume of dihydrogen evolved.

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