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
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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
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