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1.67 g mixture of Al and Zn was complete...

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.

Text Solution

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`Al+3HClrarrAlCl_(3)+(3)/(2) and Zn +2HCl rarr ZnCl_(2)+H_(2)`. Now, from 27 g Al we get 3g hydrogen so from 'x' gram
Al we will get `(x)/(9)` gram `H_(2)`. Also from 65 g Zn we get 2g hydrogen, so, from `(1.67-x)` gram Zn we will get :
`(2)/(65)(1.67-x)" gram "H_(2)=(0.0513-0.0307x)" gram "H_(2).`
Total moles of hydrogen produced `=(1.69)/(22.4)xx2=0.07544xx2=0.1508g`
Now, finally`" "0.1508=(0.0513-0.0307x)+0.1111x=0.0513+0.0804x`
`0.0804x=0.0995 rArr x=1.237g`
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