Question

If 22.8 g of nitrogen with 4.93 g of hydrogen are combined in a closed system, what mass of ammonia will be produced?

Answer 1

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