1.0 g an alloy of aluminium and magnesium when treated with excess of dilute HCl forms magnesium chloride and aluminium chloride and hydrogen collected over mercury at 0^(@)C has a volume of 1.20 L at 0.92 atmospheric pressure. Calculate the composition of the alloy.

1.0 g an alloy of aluminium and magnesium when treated with excess of

1.	1.0 g an alloy of aluminium and magnesium when treated with excess of dilute HCl forms magnesium chloride and aluminium chloride and hydrogen collected over mercury at 0^(@)C has a volume of 1.20 L at 0.92 atmospheric pressure. Calculate the composition of the alloy.
Answer» Solution :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 volumeof `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.`