Part A How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2?

Question

Part A How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. 117 mol NH3 SubmitHintsMy AnswersGive UpReview Part Incorrect; Try Again; 5 attempts remaining Part B How many grams of NH3 can be produced from 3.53 mol of N2 and excess H2. Express your answer numerically in grams. 5 g NH3 SubmitHintsMy AnswersGive UpReview Part Incorrect; Try Again; 5 attempts remaining Part C How many grams of H2 are needed to produce 14.77 g of NH3? Express your answer numerically in grams. g H2 SubmitHintsMy AnswersGive UpReview Part Part D How many molecules (not moles) of NH3 are produced from 6.44×104 g of H2? Express your answer numerically as the number of molecules. molecules

Explanation / Answer

N2 + 3H2 ----> 2NH3

1 mol N2 = 3 mol H2 = 2 mol NH3

SO that,

part A

No of mol of NH3 produced = 12*2/3 = 8 mol

part B

No of mol of NH3 produced = 3.53/2 = 1.765 mol

mass of NH3 produced = 1.765*17 = 30 grams

part c

No of mol of NH3 produced = 14.77/17 = 0.87 mol

NO of mol of H2 required = 0.87*3/2 = 1.305 mol

mass of H2 required = 1.305*2 = 2.61 grams

part D

NO of mol of H2 = (6.44*10^-4)/2 = 0.000322 mol

No of molecule of NH3 produced = 0.000322*2/3*6.023*10^23 =

   =1.293*10^20 molecules

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