Constants Periodic Table 1.78 g H2 is allowed to react with 9.77 g N2, producing
ID: 711956 • Letter: C
Question
Constants Periodic Table 1.78 g H2 is allowed to react with 9.77 g N2, producing 1.07 g NH3 The Haber-Bosch process is a very important industrial process. In the Haber-Bosch process hydrogen gas reacts with nitrogen gas to produce ammonia according to the equation Part A 3H2 (g) N2 (g) 2NH3 (g) What is the theoretical yield in grams for this reaction under the given conditions? The ammonia produced in the Haber-Bosch process has a wide range of uses, from fertilizer to pharmaceuticals. However, the production of ammonia is difficult, resulting in lower yields than those predicted from the chemical equation Express your answer to three significant figures and include the appropriate units View Available Hint(s) ValueUnits Submit Part B What is the percent yield for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units View Available Hint(s) ValueUnits SubmitExplanation / Answer
Number of moles of H2 = 1.78g / 2.016 g/mol = 0.883 mole
Number of moles of N2 = 9.77 g / 28.013 g/mol = 0.349 mole
from the balanced equation we can say that
3 mole of H2 requires 1 mole of N2 so
0.883 mole of H2 will require
= 0.883 mole of H2 *(1 mole of N2 / 3 mole of H2)
= 0.294 mole of N2 is required
but we have 0.349 mmole of N2 which is in excess so
H2 is the limiting reactant
from the balanced equation we can say that
3 mole of H2 produces 2 mole of NH3 so
0.883 mole of H2 will produce
= 0.883 mole of H2 *( 2 mole of NH3 / 3 mole of H2)
= 0.589 mole of NH3
mass of 1 mole of NH3 = 17.031 g
so the mass of 0.589 mole of NH3 = 10.0 g
Therefore, theoretical yield of NH3 = 10.0 g
percent yield = (actual yield / theoretical yield)*100
percent yield = (1.07 / 10.0)*100 = 10.7
Therefore, percent yield = 10.7 %