How many grams of nh3 can be produced from 2.27mol of n2 and excess h2.

Ok so you have 2.27 moles of N2 react with excess H2 to form NH3.
N2 + H2 -> NH3

2.27 moles + excess -> x moles

If we look at the ratio between N2 and NH3 with see that the ratio is 1 : 1
(as the number infront N2 and NH3 is 1). Now find the R.M.M (Relative atomic mass) of NH3 i.e
N- 14
H- 1 * 3 -> 3
NH3 - 17
After this multiply the R.M.M by the molarity 17 * 2.27 = 38.59 grams
when you have 4.54 do the exact same thing. 17 * 4.54 = 77.18 grams
Hope this helps :).

It is a product