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)) to 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 - 4 xx 0.2 = 0.2`