Saved Bookmarks
| 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)`A. `10.2 gm`B. `51 gm`C. `8.5 gm`D. `0.6 gm` |
|
Answer» Correct Answer - A `{:(,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` |
|