If `5gH_(2)` is mixed with 14 g of `N_(2)` for the following reaction.
`N_(2)+3H_(2)to2NH_(3)`
At the end mass of `H_(2)` left unreacted is -g.

Answer

`2`

Explanation

Number of moles of 5g`H_(2)`
`(N_(H_(2)))=5/2=2.5`mol
Number of moles of `N_(2)`
`(N_(N_(2)))=14/28=0.5`mol
`N_(2)+3H_(2)to2NH_(3)`
1 mole of `N_(2)` reacts with 3 moles of
`H_(2)` 0.5 mole of `N_(2)` reacts with
1.5 moles of `H_(2)`
`impliesn_(H_(2))` unreached
`=2.5-1.5=1` mol
1 mol `H_(2)=2gH_(2)`