Question

How many grams of NH3 can be produced from 4.89 mol of N2 and excess H2.

Answer 1

166.56g of NH3 are produced

Step-by-step explanation:

The reaction of N2 with H2 to produce NH3 is:

N2 + 3H2 → 2NH3

Where 1 mole of N2 produce 2 moles of NH3

4.89 moles of N2 with excess of H2 produce:

4.89 moles N2 * (2 moles NH3 / 1 mole N2) = 9.78 moles NH3

As molar mass of NH3 is 17.031g/mol, mass of NH3 produced is:

9.78 moles * (17.031g / mol) =

166.56g of NH3 are produced