Question

Zinc and magnesium metal each reacts with hydrochloric acid to make chloride salts of the respective metals, and hydrogen gas. A 12.50-g mixture of zinc and magnesium produces 0.6754 g of hydrogen gas upon being mixed with an excess of hydrochloric acid. Determine the percent magnesium by mass in the original mixture.

Answer 1

the reaction of Zinc and magnesium with hydrogen is

Zn+ 2HCl -------->ZnCl2+ H2(g) and Mg+ 2HCl----------->MgCl2 + H2

1 mole of Zn reacts with HCl to give 1 mole of H2 as well as 1 mole of Mg reacts with HCl to also give 1mole of H2

let x= mass of Zinc in the mixture, mass of Mg in the mixture = 12.5-x

moles = mass/atomic weight, atomic weigts ( g/mole) : Zn = 65.38 , Mg= 24

moles : Zinc= x/65.3 Mg= (12.5-x)/24

moles of H2 produced due to reaction of Zinc with HCl = x/65.3 and Mg with HCl= (12.5-x)/24

total moles of H2 prooduced =   x/65.3+(12.5-x)/24=0.015x+(12.5-x)*0.042=0.015x-0.042x+0.525=0.525-0.027x

mass of Hydrogen produced = moles of hydrogen* molar mass = 2*(0.525-0.027x)= 1.05-0.054x

mass of hydrogen produced = 0.6754 gm

1.05-0.054x =0.6754

0.054x= 1.05-0.6754

x= 6.93 gm mass of zinv in the mixture = 6.93 gm, mass of Mg = 12.5-6.93= 5.57 gm

mass % of Mg in the mixture = 100*5.57/12.5=44.56%