Chemistry, asked by afreennisha30, 1 year ago

MOLE CONCEPT
One gram of an alloy of aluminium and magnesium when heated with excess of dil. HCl forms magnesium chloride,
aluminium chloride and hydrogen. The evolved hydrogen collected over mercury at 0 C has a volume of 1.2 litres
at 0.92 at pressure. Calculate the composition of the alloy.​

Answers

Answered by rkyadavqa
7

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

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