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

(a). Only Al gives `H_2`, not the other metal below `H_2`.
`2Al+6"HCl"to2AlCl_3+3H_2`
Ew of `H_2=" Eq of "Al (xg)`
`(0.672)/(11.2)=((x)/(27))/(3)`
`(1 " mol of "H_2=2 " Eq of "H_2=2.4L)`
`(1 " Eq of "H_2=11.2L)`
`thereforex=0.54 of Al`
The weight of unknown metal `(M)=1.82-0.54=1.28g`
(b). `4Al+3O_2to2Al_2O_3`
`{:(4 mol Al, 3 " mol of "O_2),(4xx27g,3xx32g O_2):}`
`4xx27g of Al implies3xx32g O_2`
`0.54 of "Al "implies(3xx32)/(4xx27)xx0.54 g of O_2`
`(1 " mol of "O_2=32g=22.4L)`
Volume of `O_2=(22.4)/(32)xx(3)/(4)xx(32)/(27)xx0.54`
`=0.336 L of O_2`
`O_2` used for metal `M=(0.56-0.336)L=0.224L`
Let M be monovalent.
`1 " mol of "M= 1 " mol of "O_2`
`implies(1.28)/("Atomic weight")=(0.224)/(22.4)" mol of "O_2`
`therefore` Atomic weight`=(1.28)/(0.224)xx22.4`
`=128` which is not possible
So the metal is divelent
`2 mol=1 " mol of "O_2`
`therefore` Atomic weight `=(128)/(2)=64`
i.e., metal is copper.