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6 g of H(2) reacts with 14 g N(2) to ...

6 g of `H_(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 ……

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The correct Answer is:
3
`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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