How many grams of nh3 can be produced from 3.80 mol of n2 and excess h2?

The  balanced equation :N2 + 3H2 --> 2NH3

1) Let's find the molar mass of NH3 because it will be needed later.
 To determine the molar mass you have to consult the periodic table and see the atomic mass of each element. If the element is repeted, for example 3 times, you multiple the atomic mass per 3.
NH3= 14 + 1 x 3= 17 g/mol

2) By looking at the balanced equation, you can see that 2 mol of NH₃ were produced with only one mol of N2. How much NH3 will result when we have 3,80 mol of N2?
So now we solve it with a porportion:

2 mol NH31 mol of N2
x               3.80 of N2

x= 2 x3,8 = 7,6mol NH3

3)  the only thing left to do is convert the moles obtained, to grams. We can use the formula:
nº of moles x molar mass= mass
7,6mol NH3  x 17 g/mol= mass of NH3
mass= 129,2 grams of NH3

Lowest form: