Question

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

3H2(g)+N2(g)→2NH3(g)3H2(g)+N2(g)→2NH3(g)

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.

1.36 g H2 is allowed to react with 9.75 g N2, producing 1.75 g NH3

What is the theoretical yield in grams for this reaction under the given conditions?

What is the percent yield for this reaction under the given conditions?

Answer 1

Augwert 3 H219) + N219) mores of M2 = mars mars molar mare 2 HH319) = 1.36g, = 0.68 mo) - 2gmont moter of H2 = man molar manz = 9.75 g = 0.348 mol New 3 mole of H₂ needs 1 mol of N2 0.68 mol of H₂ needs 1 ma of N2 x 0.68 molte than 3 mol H2 COPA OOOL 0.23 mol of N2 meleg of N₂ we have a 0.348 mol (more) .. M2 is limiting reactant moter of Alla prodered = 2 x meleg of H₂ Cangermed CHO O = = 2 x 0.08 mo) = 0.453 mo) mans of NM3 prodened = mol x molar mare = 0.963 mes xl7gmarl @ Theoretical yield = 7.70g. t. yield = 1.759 x 100% = 22 Theo X 100 . = 22.7 y. langweil