The excess reactant would be H2 by 6.8 g.
N2 and H2 react according to the following balanced equation:
[tex]N_2 +3H_2 --->2 NH_3[/tex]
The mole ratio of N2 and H2 is 1:3. For every 1 mole of N2. 3 moles of H2 would be needed for complete reaction.
If, mole = mass/molar mass
mole of 15.0 g of N2 = 15/28
= 0.536 moles
Mole of 10.0 g of H2 = 10/2
= 5 moles
Stoichiometrically, the mole of H2 required for 0.536 mole of N2 should be:
0.536 x 3 = 1.6 moles.
Thus, the excess mole of H2 would be:
5 - 1.6 = 3.4 moles
Mass of excess moles of H2 = 3.4 x 2
= 6.8 g.
More on calculating moles can be found here: https://brainly.com/question/21085277
