One gram of an alloy of `Al and Mg` when treated with excess of dilute `HCI` forms `MgCI_2, AICI_3` and hydrogen. The evolved hydrogen collected over Hg at 0°C has a volume of `1.20` litres at `0.92` atm pressure. Calculate the composition of the alloy. `("AI" = 27, "Mg" = 24)`

Explanation

The equations, `{:(Ai+3h^(+)" "to Ai^(3+) + (3)/(2) H_(2)),(Mg + 2H^(+) to Mg^(2+) + H_(2) ):}`
show that 1 mole of Al produces `(3)/(2)` moles of hydrogen and
1 mole of Mg produces 1 mole of hydrogen.
Thus the mole equation is ,
`(3)/(2)` moles of Al + 1 mole of Mg = 1 mole of `H_(2)`
Let the weight of Al be x g
`:. ` wt. of Mg = (1 - x) g
`(3)/(2) xx (x)/( 27) + (1 - x)/( 24) = ("vol. of " H_(2) " at NTP")/( 22. 4)`
Volume of `H_(2)` at NTP ` = (1 . 2 xx 0 . 92)/( 273) xx (273)/( 1) = 1 . 1 .104` liters
(using `(p_(1)V_(1))/(T_(1)) = (p_(2)V_(2))/(T_(2)))`
`:. (3)/(2) xx (x)/(2) + (1 - x)/( 24) = (1 . 104)/( 22.4) , x = 0 . 55`
Wt. of Al = 0 . 55 g, wt. of Mg 0 . 45
Thus `%` of Al `= ( 0 . 55)/( 1) xx 100 = 55 % `
and `% " of Mg " = (0.45)/( 1) xx 100 = 45%`