`1 g` of an alloy of `Al` and `Mg` reacts with excess `HCl` to form `AlCl_(3)`, `MgCl_(2)`, and `H_(2)`. The evolved `H_(2)` collected over mercury at `0^(@)C` occupied `1200 mL` at `699 mm Hg`. What is the composition of alloy?

Explanation

`Al+3HCltoAlCl_(3)+(3)/(2)H_(2)`
`Mg+2HCltoMgCl_(2)+H_(2)`
Let the mass of `Al` be `x g`. Then the mass `Mg` is `(1-x)g`.
Also moles of `H_(2)=(PV)/(RT)=((699)/(760)xx1.2)/(0.0821xx273)=0.0492`
`implies ` Moles `H_(2)=(3)/(2)xx(x)/(27)+((1-x))/(24)xx1=0.0492`
`0.0555x+0.042(1-x)=0.0492`
`implies 0.0135x=7.24xx10^(-3)`
`implies x=0.536 g`
`implies %Al=53.6%`, `%Mg=46.4%`