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An alloy of Al and Mg weighing 1.26 g re...

 An alloy of Al and Mg weighing 1.26 g required 0.448 L of `N_(2)` gas at NTP. Calculate mol% of Al and Mg in the alloy.

Text Solution

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Both Al and Mg react with `N_(2)`
Let Al=xg, Mg (1:26 - x)g: N2 used 0.448 Lat NTP = 0,56 g
`overset(2xx27)g(2AI)+overset(28g)(N_(2))to2AIN`,   For x g AI=`(28x)/(54)=(14x)/(27)gN_(2)`
=`(7)/(18)(1.26-x)gN_(2)`
`because` Total `N_(2)` = 0.56 g (given)
`(14x)/(27)+(7)/(18)(1.26-x)=0.58g`
`therefore` .x = 0.54 g = 0.02 mol Al, (1.26 - x) = 0.72 g = 0.03 mol Mg
`therefore`  .Al = 40%, Mg = 60 %
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