Question

1.13 g H, is allowed to react with 9.69 g N2, producing 2.19 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 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. UA ? Value Units Submit Request Answer 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. КА ? Value Units

Answer 1

1.13X 4 : 66 67 g N2 consumed so A A in IS the stochometry 6 NH3 Ig 12 gedure Thea retical Yield of NH3 grams ✓ amount of NH3 Produced according to will 100% icon version.'." N2 + 3H2 h produce 2 mol in NH₂O N D considering molar mass y buenosat 1289 N₂ react with dg Hz produce 34 g NH Gg biz react with 289 N2 so According to stocheometric canation I mol Nå react with 3 mol H2 g and 3 tn 28 2.6667 g Na sol 13 g H₂. & eact with - g. 273339 N2 So H₂ is completely it and Nam is limiting reagent bord is present in excess: 1 According to stochiometall eauotion 68 Н, 28 349 NH3 so 1.13 g Hz produce 34 X1:13 9 NHL 6 6.40 3.3 g NH3 theorence yield anon so ng ram's = 16.40 I NHz 1/ percent Yield X100 % of NH₃ = Actual Yield Theoretical Yield Yield of Nidz 6:4 of NH3 = 2.19 g X100 = 34.2% 6.40 6.409 Theoretical Actual Yield percentajie' yceed"=" 2:14 so percentage Yield is 34.2% Please Like if it was helpfull. Thanks Scanned with CamScanner