1.67 g mixture of Al and Zn was completely dissolved in acid and evolved 1.69 L of `H_2` at STP. Calculate the weight Al and Zn in the mixture.

Explanation

Let `a` and `b g` be weight of `Al` and `Zn` in mixture
`a+b= 1.67`….(1)
`Al+3HClrarrAlCl_(3)+(3//2)H_(2)`,
`Zn+2HClrarrZnCl_(2)+H_(2)`
`:' 27 gAl` gives `(3//2)xx22.4 "litre"H_(2)`
`( :' at.wt. of Al= 27)`
`:. agAl` gives `(3xx22.4xxa)/(2xx27)"litre"H_(2)`
Similarly, `65 g Zn` gives `22.4 "litre" H_(2)`
`( :' at.wt. of Zn=65)`
`:. bgZn` gives `(22.4xxb)65"litre"H_(2)`
`:. (3xx22.4xxa)/(2xx27)+(22.4xxb)/(65)=1.69`.....(2)
By eq. (1) and (2), `a= 1.25 g, b=0.42 g`