Question

PLZ HELP IM RUNNING OUT OF TIME!!!!!

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. However, you calculate you should have recovered 1.33 moles NH3. What is the percent yield of the reaction?
Question options:
(a)60%
(B) 27%
(C)41%
(D)246%

Answer 1

I think its, C) 41%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 "C" to be considered correct..

Sorry if its wrong!