Question

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l). In an industrial synthesis of urea, a chemist combines 136.4 kg of ammonia with 211.4 kg of carbon dioxide and obtains 168.4 kg of urea. Determine the theoretical yield of urea, limiting reactant, and percent yield for the reaction.

Answer 1

Hy mass = 136.4kg = 136400 gr 2 Nitz + co - CHNO + HO NH: Molasmass = 12 glmol co Mass=211.4kg = 211400 gr co: Molar mass 244 glmal CH No Molarmass = 60 gimol I From equation: Imole co react with a moles NH₂ (LX 442gr co react with (2 x 172gr NH, 4488co react with 34 gr NH, 211400 qr co react with = 211400 x 34) & NH3 44 mass of N required - 163354-5 gr required But we have only 136400 gr NH so Why is limiting reactant a from equation. Emoles Why produce Imole ch No (2X 17288 NH, produce Cix Go2gr cHMO 3488 NH, produce to go CH₂ NO 136400 gr Nt produce = 136400x60)88 CHNO - 34 Theoretical field of CH NO = 240.705.88 gr actual yield of a urea = 168.4kg =168400 gr actual yield xLoo 7- yield Theoretical Xloo 168400 69 - 961". 240 205-88 Ratepless e