How many grams if nh3 can be produced from 2.57mol of n2 and excess h2?

Answer 1

2.57 mol of N2 --> 2 * 14 * 2.57 = 72 g of N2

that is 5.14 mol of N so 3*5.14 = 15.4 mol of H which is 1 g/mol

15.4 mol * 1 = 15.4 g of H
so
72 + 15.4 = 87.4 grams of NH3