To calculate the mass of NH3 produced, we need to first determine the limiting reactant. To do this, we calculate the number of moles of each reactant using their respective molar masses:

Moles of N2 = 10 g / 28 g/mol = 0.357 mol

Moles of H2 = Excess (assumed infinite)

The balanced chemical equation for the reaction is:

N2 + 3H2 -> 2NH3

From the stoichiometry of the equation, we see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Therefore, N2 is the limiting reactant in this case.

Now we can calculate the moles of NH3 produced:

Moles of NH3 = 0.357 mol N2 * (2 moles NH3 / 1 mole N2) = 0.714 mol NH3

Finally, we calculate the mass of NH3 produced using its molar mass:

Mass of NH3 = 0.714 mol * 17 g/mol = 12.14 g

Therefore, 12.14 g of NH3 is produced from the reaction of 10 g N2 with excess H2.