Question

$1\cdot 0g$ an alloy of aluminum and magnesium when treated with an excess of dilute $HCl$ form magnesium chloride and aluminum chloride and hydrogen collected over mercury at $0^{\circ }$ has a volume of $1\cdot 20L$ at $0\cdot 92$ atmospheric pressure . Calculate the composition of the alloy.

Answer 1

Suppose AI in the alloy = $xg$
Then Mg in the alloy $=(1-x)g$
AI and Mg in the alloy will react with HCI acid as follows:
$2AI$ $+6HCI\to 2AIC{I}_3+$ $3H_2$
$2\ast 27=54g$ $3\ast 22\cdot 4L$ at S.T.P.
$Mg$ $+2HCI\to MgC{I}_2+$ $H_2$
$248$ $22\cdot 4L$ at S.T.P.
$H_2$ produced from $xg$ of $AI=\frac{3\ast 22\cdot 4}{54}\ast x$
$=\frac{22\cdot 4x}{18}L$ at S.T.P.
$H_2$ produced from $(1-x)g$ of $Mg=\frac{22\cdot 4}{24}\ast x$
$=\frac{22\cdot 4}{24}\ast (1-x)L$ at S.T.P.
$\therefore$ Total $H_2$ produced at S.T.P
$=\frac{22\cdot 4x}{18}+\frac{22\cdot 4(1-x)}{24}L$ at S.T.P.
Let us now convert the actual volume of $H_2$ produced to volume at S.T.P.
$P_1=0\cdot 92atm$ $P_2=1atm$
$V_1=1\cdot 20L$ $V_2=?$
$T_1=273K$ $T_2=273K$
$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}\therefore \frac{0\cdot 92\ast 1\cdot 20}{273}=\frac{1\ast V_2}{273}$

or $V_2=1\cdot 104L$

$\therefore \frac{22\cdot 4x}{18}+\frac{22\cdot 4(1-x)}{m}=1\cdot 104$

or $4\ast 22\cdot 4x+3\ast 22\cdot 4(1-x)=1\cdot 104\ast 72$
or $89\cdot 6x+67\cdot 2-67\cdot 2x=79\cdot 488$

or $22\cdot 4x=12\cdot 288$ or $x=0.5486g$

$\therefore$ % of $Al=54.86$
and % of Mg $=100-54\cdot 86=49\cdot 14$