1.0g of an alloy of magnesium and aluminium dissolves exactly in 100ml of 1.0 M HCl. The percentage composition of the alloy is

Answer:

The number of moles of hydrogen are n=

RT

PV

=

0.0821×273

0.92×1.2

=0.0493

Mg+2HCl→MgCl

2

+H

2

2Al+6HCl→2AlCl

3

+3H

2

Let x g of Mg and 1-x g of Al are present.

This corresponds to

24.3

1−x

moles of Mg and

27

x

moles of Al.

The number of moles of hydrogen formed are

24.3

1−x

+

18

x

.

But this is equal to 0.0493 moles.

24.3

1−x

+

18

x

=0.0493

18−18x+24.3x=21.57

6.3x=3.57

x=0.547g= mass of Al

Mas of Mg =1−0.547=0.454g

Thanks