1.

If following reaction is started with 6 gm of H_(2) and 14 gm of N_(2) then mass of NH_(3) formed will be : N_(2) (g) + 3H_(2) (g) overset(60%)rarr 2NH_(3) (g)

Answer»

`10.2 gm`
`51 gm`
`8.5 gm`
`0.6 gm`

Solution :`{:(,N_(2) (G),+,3H_(2) (g),OVERSET(60%)rarr,2NH_(3) (g),),(Moles,(1)/(2),,3,,,),(N_(2) "is LR",,,,,,):}`
Moles of `NH_(3)` PRODUCED `= (1)/(2) xx 2 xx 0.6 = 0.6` mole
Mass of `NH_(3) = 0.6 xx 17 = 10.2 gm`


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