Answer:

The % of Al in the Alloy is 54.91%

The % of Mg in the Alloy is 45.09%

Explanation:

Given - "One gram of an alloy of aluminium and magnesium when heated with excess of dil. HCl forms magnesium chloride, aluminium chloride and hydrogen."

So let`s write equations as per given equations;

Al + HCl     ⇄   ALCL3 + H2

Let`s balance the equation;

2Al + 6HCl     ⇄     2AlCl3 +   3H2

From the equation - 2 mole of Al produces 3 mole of H2

⇒ 2* 27 gm of Al produces 3 moles of H2

At STP 1 mole H2 gas is equal to 22.4 litres of gas

⇒ 54 gm of Al produces 3*22.4 litres of H2 gas at STP......EQ 1

Let`s calculate for Mg

Mg(s) + 2HCl (dil.)    ⇄   MgCl2(s)    + H2

1 mole of Mg produces 1 mole of H2 gas

24 gm of Mg produces 22.4 litres of H2 gas at STP......Eq. 2

First we need to calculate volume of H2 gas at STP, ⇒ at 273 K and 1 bar

Given "The evolved hydrogen collected over mercury at 0 C has a volume of 1.2 litres at 0.92 at pressure."

From ideal gas equations - P1V1/T1  =  P2V2/T2......Eq. 3

V1 = 1.2 litres P1 = 0.92 bar and T1 = 0+273 = 273 K

V2 = ? P2 = 1 bar and T2 = 273 K

By PUTTING above values in eq. 3,

(0.92)*(1.2)/273    =     V2*(1)/273

⇒ V2 = 1.104 litres

Let`s consider the amount of Al in alloy is x gm

So, the amount of Mg in Alloy will be 1-x gm (∵ Alloy is 1 gm, given)

From Eq. 1, ∵ 54 gm Al produces 3*22.4 litres of H2 gas

∴ x gm of Al will produce 3*22.4*x/54 litres of H2 gas = 1.24x.....Eq. 4

From Eq. 2 ∵ 24 gm Mg produces 22.4 litres of H2 gas

∴ 1-x gm of Mg will produce 22.4*(1-x)/24 litres of H2 gas = 0.93(1-x) ....Eq. 5

From Eq. 4 & 5, total volume of H2 gas = 1.24x + 0.93(1-x) = 0.93 + 0.31x ....Eq. 6

The volume of H2 from Eq. 6 should be equal to evolved H2 gas at STP (V2)

⇒ 0.93   + 0.31x   =   1.104

⇒   x  = 0.5491

and 1-x would be 0.4509

% of Al = x*100/1  = 54.91

% of Mg = (1-x)*100/1 = 45.09