In the reaction, `4NH_(3)(g)+5O_(2)(g) rarr 4NO(g)+6H_(2)O(g)`, when 1 mole of ammonia and 1 mole of `O_(2)` are made to react to completion

Options

A. 1.0 mole of `H_(2)O` is produced
B. 1.0 mole of NO will be produced
C. All the oxygen will be consumed
D. All the ammonia will be consumed

Answer

C

Explanation

`{:(,4NH_(3(g)),+,5O_(2(g)),rarr,4NO_((g)),+,6H_(2)O_((g))),(t=0,1,,1,,0,,0),(t=t,1-4x,,1-5x,,4x,,6x):}`
Oxygen is limiting reagent
So, `X=1/5=0.2` all oxygen consumed
Left `NH_(3)=1-4xx0.2=0.2`