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1 mol of `N_(2)` and 2 mol of `H_(2)` are allowed to react in a 1 `dm^(3)` vessel. At equilibrium, `0.8` mol of `NH_(3)` is formed. The concentration of `H_(2)` in the vessel is
A. `0.6` mole
B. `0.8` mole
C. `0.2` mole
D. `0.4` mole

1 Answer

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Best answer
Correct Answer - B
`N_(2)+3H_(2)hArr2NH_(4)`
`{:(t=0,1"mole",2"mole",,0),(t=eq,1-x,2-3x,,2x=0.8):}`
`x=0.4`
mole of `N_(2)=0.6`
mole of `H_(2)=0.8`

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