Question

For the reaction of ammonia with molecular oxygen, forming nitrogen monoxide and water. How do you calculate the theoretical yield of `NO` when 17.0g `NH_3` reacts with 16.0g `O_2`?

Answer 1

Well first you write a stoichiometrically balanced equation:

`2NH_3(aq) + 5/2O_2(g) rarr 2NO(g) +3H_2O(l)`

Explanation:

Given stoichiometric oxygen, ammonia gives an equimolar quantity of `NO`.

`"Moles of ammonia"` `=` `(17.0*g)/(17.03*g)` `~=` `1*mol`

`"Moles of dioxygen"` `=` `(16.0*g)/(32.03*g)` `~=` `0.50*mol`

Clearly, dioxygen is in deficiency, and only `4/5xx0.5*mol=0.40*mol` `NH_3` will react.

And thus only `0.40*mol` `NO` will be produced, i.e. `0.40*molxx30.01*g*mol^-1` `=` `12.0*g` `NO`.