Question

One gram of an alloy of aluminium and magnesium when heated with excess of dilute $HCl$ forms magnesium chloride, aluminium chloride and hydrogen. The evolved hydrogen collected over mercury at $0^{o}C$ has volume of $1.2$ litres at $0.92$ atm pressure. Calculate the composition of the alloy.

• A. $Al=0.546$ g
• B. $Al=0.454$ g
• C. $Mg=0.454$ g
• D. $Mg=0.546$ g

Answer 1

Answer: (A), (C)

$Mg=0.454$ g
$Al=0.546$ g

The number of moles of hydrogen are $n=\frac{PV}{RT}=\frac{0.92\times 1.2}{0.0821\times 273}=0.0493$
$Mg+2HCl\to MgC{l}_2+H_2$
$2Al+6HCl\to 2AlC{l}_3+3H_2$
Let x g of Mg and 1-x g of Al are present.
This corresponds to $\frac{1-x}{24.3}$ moles of Mg and $\frac{x}{27}$ moles of Al.
The number of moles of hydrogen formed are $\frac{1-x}{24.3}+\frac{x}{18}$.
But this is equal to 0.0493 moles.
$\frac{1-x}{24.3}+\frac{x}{18}=0.0493$
$18-18x+24.3x=21.57$
$6.3x=3.57$
$x=0.547g=$ mass of Al
Mas of Mg $=1-0.547=0.454g$