One gram of an alloy of aluminium and magnesium when heated with excess of dil. `HCI` forms magnesium chloride, aluminium chloride and hydrogen. The evolved hydrogen collected over mercury at `0^(0)C` has a volume of `1.2` litre at `0.92 atm` pressure. Calculate the composition of the alloy.

Explanation

Let us first calculate the volume of `H_(2)` at N.T.P.
Applying
`(p_(1)V_(1))/T_(1) =(p_(2)V_(2))/T_(2)`
`p_(1) = 0.92 atm, p_(2) = 1 atm`
`V_(1)=1.20 L, V_(2)=?`
`T_(1) = 273 K , T_(2) = 273 K`
`therefore V_(2)=(p_(1)V_(1)T_(2))/(p_(2)T_(1)) = (0.92 xx 1.20 xx 273)/(1 xx 273)`
= 1.104 L
Let mass of Al in alloy =xg
Mass of Mg in alloy =(1-x) g
`underset(54 g)(2Al) + 6HCl to 2AlCl_(3) + underset(67.2 L)(3H-(2))`
`underset(24 g)(Mg) + 2HCl to MgCl_(2) + underset(22.4 L)(H_(2))`
54 g of Al give `H_(2)` at N.T.P. = 67.2 L
xg of Al give `H_(2)` at NTP `=(67.2 xx x)/54`
Similarly, 24 g of Mg give `H_(2)` at NTP= 23.4 L
(1-x) g of Mg will give `H_(2)` at N.T.P. `=(22.4)/24 xx (1-x)`
Total `H_(2)` liberated
`(67.2 x)/54 + (22.4 (1-x))/24 = 1.104`
Solving for x, we get
x=0.5486 g
`therefore` Mass of Al in the alloy = 0.5486 g
% of Mg in alloy =100 - 54.86 = 54.86
% of Mg in alloy = 100 - 54.86 = 45.14