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6 g ofH_(2) reacts with 14 g N_(2)to form NH_(3)till the reaction completely consumes the limiting reagent. The mass of other reactant (in g) left are ……

Answer»


Solution :`3H_(2) + N_(2) RARR 2NH_(3)`
GIVEN moles : `(6)/(2) = 3 ""(14)/(28) = 0.5`
3 mol of`H_(2)`REQUIRED 1 mol of`N_(2)`so `N_(2)`is the limiting REAGENT. 0.5 mol of `N_(2)`.will REACT with 1.5 mol`H_(2)`. So 1.5 mol`H_(2)`will be left.
Mass of 1.5 mol`H_(2) = 2 xx 1.5 g = 3g`


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