1 . 0 g of an alloy of Aland Mg when treated withexcess of dil. HCI forms MgCl_(2), AlCl_(3) and hydrogen. Theevolved hydrogen,collected over Hg at 0^(@) C has a volume of 1.2 litters at 0.92 atm pressure. Calculate the composition ofthe alloy. (al = 27 and Mg = 24)

1 . 0 g of an alloy of Aland Mg when treated withexcess of dil. HCI fo

1.	1 . 0 g of an alloy of Aland Mg when treated withexcess of dil. HCI forms MgCl_(2), AlCl_(3) and hydrogen. Theevolved hydrogen,collected over Hg at 0^(@) C has a volume of 1.2 litters at 0.92 atm pressure. Calculate the composition ofthe alloy. (al = 27 and Mg = 24)
Answer» Solution :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 hydrogenand 1 moleof Mg produces 1 mole of hydrogen. Thusthe moleequation 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)` Volumeof `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 Mg0 . 45 Thus`%` of Al `= ( 0 . 55)/( 1) xx 100 = 55 % ` and `%" of Mg " = (0.45)/( 1) xx 100 = 45%`