Question

Ammonia gas is synthesized according to the balanced equation below. N2(g) + 3 H2(g) -> 2 NH3(g) If 2.50 L N2 react with 7.00 L H2, what is the theoretical yield (in liters) of NH3? Assume that the volumes of reactants and products are measured at the same temperature and pressure.

Answer 1

N₂(g) + 3 H₂(g) -> 2 NH₃(g)

Volume of N₂ : 2.5 L

Moles of N₂: 2.5 L / 22.4 L = 0.112 mol

One mole of an ideal gas will occupy a volume of 22.4 liters at STP.

Volume of H₂ : 7 L

Moles of H₂: 7 L / 22.4 L = 0.3125 mol

Limiting reagent :H₂

Excess Reagent: N₂

Stoichimetry of the reaction is 3:1 (H₂ to N₂)

0.3125 moles of H₂ react with (0.3125/3) 0.104 moles of N₂ to produce (0.104x2) 0.208 moles of ammonia

So the theoretical yield of ammonia is 0.208 moles = 0.208 x 22.4L = 4.66 L