One gram of an alloy of aluminium and magnesium when heated with excess of dil. `HCI` forms magnesium chloride, aluminium chloride and hydrogen. The evolved hydrogen collected over mercury at `0^(0)C` has a volume of `1.2` litre at `0.92 atm` pressure. Calculate the composition of the alloy.

Explanation

Suppose Al in the alloy = x g
Then Mg in the alloy `=(1-x)g`
Al and Mg in the alloy will react with HCl acid as follows :
`{:(underset(2xx27=54g)(2Al)+6HCl rarr2AlCl_(3)+underset(3xx22.4"L at S.T.P.")(3H_(2))),(" "underset(24g)(Mg)+2HCl rarr MgCl_(2)+underset("22.4 L at S.T.P.")(H_(2))):}`
`H_(2)" priduced from x g of Al"=(3xx22.4)/(54)xx x=(22.4x)/(18)"L at S.T.P.`
`H_(2)" produced from "(1-x)g" of Mg"=(22.4)/(24)xx(1-x)"L at S.T.P."`
`therefore" Total "H_(2)" produced at S.T.P."=(22.4x)/(18)+(22.4(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)=0.92atm" "P_(2)=1atm),(V_(1)=1.20L" "V_(2)=?),(T_(1)=273K" "T_(2)=273K),((P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)) therefore (0.92xx1.20)/(273)or v_(2)=1.104L):}`
`"or "4xx22.4x+3xx22.4(1-x)=1.104xx72`
`"or "89.6x+67.2-67.2x=79.488`
`"or "22.4x=12.888 or x=0.5486g`
`therefore" "%" of Al"=54.86 and %" of Mg"=100-54.86=49.14.`