limiting reactant and therotical yeild of 0.528 moles N2, 1.28 moles H2

Answer 1

I presume we're making ammonia, N₂ + 3 H₂ --> 2 NH₃.

(0.528 mole N₂) * (2 mole NH₃ /1 mole N₂) = 1.05₆ mole NH₃.

(1.28 mole H₂) * (2 mole NH₃ /3 mole H₂) = 0.853 mole NH₃.

Hydrogen is the limiting reactant, and (0.853 mole NH₃) * (17.03 g NH₃ /1 mole NH₃) = 14.5 g NH₃ for the theoretical yield.