Question

How many grams of ammonia can be prepared from 77.3 g of nitrogen and 14.2 g of hydrogen? Write the balanced equation, too.​ Please provide a brief explanation in sentences as well.

Answer 1

The balanced equation is

N2 + 3H2 -----> 2NH3

number of moles of N2 = 77.3 g / 28.0 g/mol = 2.76 mole

number of moles of H2 = 14.2g / 2.016 g/mol = 7.04 mole

from the balanced equation we can say that

1 mole of N2 requires 3 mole of H2 so

2.76 mole of N2 will require 8.28 mole of H2

But we have only 7.04 mole of h2 so H2 is the limiting reactant

from the balanced equation we can say that

3 mole of H2 produces 2 mole of NH3 so

7.04 mole of H2 will produce 4.69 mole of NH3

1 mole of NH3 = 17.03 g

4.69 mole of NH3 = 79.9 g

Therefore, the mass of ammonia produced will be 79.9g