How many moles of NH3 can be produced from 24.0 mol of H2 and excess N2?

3H2(g) + N2(g)→2NH3(g)
How many moles of NH3 can be produced from 24.0 mol of H2 and excess N2?

Answer:

3H2(g) + N2(g)→2NH3(g)

according to balanced reaction

3 moles of H2 gives 2 moles of NH3

24.0 moles of H2 gives (24 x 2) /3 = 16

moles of NH3 formed = 16