Answer:
27.73 g of NH₃
Solution:
The Balance Chemical Reaction is as follow,
N₂ + 3 H₂ → 2 NH₃
Step 1: Find out the limiting reagent as;
According to Equation ,
28 g (1 mole) N₂ reacts with = 6.04 g (3 moles) of H₂
So,
22.8 g of N₂ will react with = X g of H₂
Solving for X,
X = (22.8 g × 6.04 g) ÷ 28 g
X = 4.91 g of H₂
It means for total utilization of 22.8 g of N₂ we require 4.91 g of H₂, but we are provided with 4.93 g of H₂. Therefore, N₂ is the limiting reagent and will control the yield of product.
Step 2: Calculate Amount of NH₃ produced as;
According to Equation ,
28 g (1 mole) N₂ produces = 34.06 g (2 moles) of NH₃
So,
22.8 g of N₂ will produce = X g of NH₃
Solving for X,
X = (22.8 g × 34.06 g) ÷ 28 g
X = 27.73 g of NH₃
