Ammonia (NH3) can be produced by the reaction of hydrogen gas with nitrogen gas:

3H2 + N2 = 2NH3

A chemist reacts 2.00 mol H2 with excess N2. The reaction yields 0.54 mol NH3. What is the percent yield of the reaction?

a) 25%

b) 40%

c) 60%

d) 80%

Question

Conner Best

Chemistry

23 December, 10:13

Ammonia (NH3) can be produced by the reaction of hydrogen gas with nitrogen gas:

3H2 + N2 = 2NH3

A chemist reacts 2.00 mol H2 with excess N2. The reaction yields 0.54 mol NH3. What is the percent yield of the reaction?

a) 25%

b) 40%

c) 60%

d) 80%

+2

Answers (1)

Know the Answer?

Jaylan Wade · Answer 1

B) 40%

The balanced equation indicates that for every 3 moles of H2 used, 2 moles of NH3 will be produced. So the reaction if it had 100% yield would produce (2.00 / 3) * 2 = 1.333333333 moles of NH3. But only 0.54 moles were produced. So the percent yield is 0.54 / 1.3333 = 0.405 = 40.5%. This is a close enough match to option "b" to be considered correct.

Ammonia (NH3) can be produced by the reaction of hydrogen gas with nitrogen gas:3H2 + N2 = 2NH3A chemist reacts 2.00 mol H2 with excess N2. The reaction yields 0.54 mol NH3. What is the percent yield of the reaction?a) 25%b) 40%c) 60%d) 80%

Ammonia (NH3) can be produced by the reaction of hydrogen gas with nitrogen gas:

3H2 + N2 = 2NH3

A chemist reacts 2.00 mol H2 with excess N2. The reaction yields 0.54 mol NH3. What is the percent yield of the reaction?

a) 25%

b) 40%

c) 60%

d) 80%