limiting reactant and theoretical yield of 2.00 g N2, 1.00 g H2

Answer 1

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

(2.00 g N₂) * (1 mole N₂ /28.01g N₂) * (2 mole NH₃ /1 mole N₂) = 0.143 mole NH₃.

(1.00 g H₂) * (1 mole H₂ /2.016g H₂) * (2 mole NH₃ /3 mole H₂) = 0.331 mole NH₃.

Nitrogen is the limiting reactant, and (0.143 mole NH₃) * (17.03 g NH₃ /1 mole NH₃) = 2.43 g NH₃ for the theoretical yield.