Question

If 15.0 g of nitrogen reacts with 15.0 g of hydrogen, 10.5 g of ammonia is

produced. What is the percent yield?

Answer 1

Percent yield = 57%

Step-by-step explanation:

Given data:

Mass of nitrogen = 15.0 g

Mass of hydrogen = 15.0 g

Mass of ammonia produced = 10.5 g

Percent yield = ?

Solution:

Chemical equation:

N₂+ 3H₂ → 2NH₃

Number of moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 15.0 g/ 2 g/mol

Number of moles = 7.5 mol

Number of moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 15.0 g/ 28 g/mol

Number of moles = 0.54 mol

Now we will compare the moles of ammonia with nitrogen and hydrogen from balance chemical equation.

N₂ : NH₃

1 : 2

0.54 : 2×0.54 = 1.08

H₂ : NH₃

3 : 2

7.5 : 2/3×7.5= 5 mol

Theoretical yield of ammonia:

Mass = number of moles × molar mass

Mass = 1.08 × 17 g/mol

Mass = 18.36 g

Percent yield:

Percent yield = (actual yield / theoretical yield) ×100

Percent yield = (10.5 g/ 18.36 g) ×100

Percent yield = 0.57 ×100

Percent yield = 57%