How many moles of NH3 can be produced from 19.5mol of H2 and excess N2

This is a simple (as opposed to limiting reagent) stoichiometry problem.

N2 + 3H2 ==> 2NH3
Use the coefficients in the balanced equation to convert mols of anything in the equation to mols of anything else in the equation.
19.5 mol H2 x (2 mol NH3/1 mol N2) = 19.5 x 2/1 = ?

13

39

13

39

13!

To find out how many moles of NH3 can be produced from 19.5 mol of H2 and excess N2, we need to use the balanced chemical equation for the reaction between H2 and N2 to form NH3.

The balanced chemical equation is:

N2 + 3H2 -> 2NH3

According to the equation, 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

Therefore, we need to determine the limiting reactant in this reaction by comparing the number of moles of each reactant to the stoichiometry of the balanced equation.

Given that there is an excess of N2, we can focus on H2. We have 19.5 mol of H2.

Since 1 mole of N2 reacts with 3 moles of H2, we can calculate the number of moles of NH3 using the mole ratio:

19.5 mol H2 * (2 mol NH3 / 3 mol H2) = 13 mol NH3

Thus, 19.5 mol of H2 can produce 13 mol of NH3 when N2 is in excess.