Constants | Periodic Table 1.32 g H2 is allowed to react with 10.4 g N2, produci
ID: 1026031 • Letter: C
Question
Constants | Periodic Table 1.32 g H2 is allowed to react with 10.4 g N2, producing 2.38 g NHs is a very important The Haber-Bosch process 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) Part A 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. What is the theoretical yield in grams for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units View Available Hint(s) alueUnits Submit Part B What is the percent yield for this reaction under the givern conditions? Express your answer to three significant figures and include the appropriate units View Available Hint(s) ? alueUnitsExplanation / Answer
N2(g) + 3H2(g) ------------> 2NH3(g)
no of moles of N2 = W/G.M.Wt
= 10.4/28 = 0.37moles
no of moles of H2 = W/G.M.Wt
= 1.32/2 = 0.66moles
1 moles of N2 react with 3 moles of H2
0.37 moles of N2 react with = 3*0.37/1 = 1.11moles of H2 is required
H2 is limiting reactant
3 moles of H2 react with N2 to gives 2 moles of NH3
0.66 moles of H2 react with N2 to gives = 2*0.66/3 = 0.44moles of NH3
mass of NH3 = no of moles * gram molar mass
= 0.44*17 = 7.48g
Theoretical yiled = 7.48g
percent yield = actual yield*100/Theoretical yiled
= 2.38*100/7.48 = 31.8% >>>>answer