1.72 g H is allowed to react with 10.0 g N2. producing 1.22 g NHs Part A What is
ID: 550340 • Letter: 1
Question
1.72 g H is allowed to react with 10.0 g N2. producing 1.22 g NHs Part A 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 Hints 6.25 Submit My Answers Give Up Incorrect; Try Again; 4 attempts remaining; no points deducted 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. Hints 25 Submit My Answers Give Up Incorrect; Try Again; 3 attempts remainingExplanation / Answer
N2 + 3H2 ---> 2 NH3
no of mol of H2 reacted = w/mwt = 1.72/2 = 0.86 mol
no of mol of N2 reacted = 10/17 = 0.6 mol
limiting reactant = H2
noof mol of NH3 formed = 0.86*2/3 = 0.573 mol
theoretical yield of NH3 = 0.573*17 = 9.74 g
practical yield of NH3 = 1.22 g
percent yield = practical yield/theoretical yield*100
= 1.22/9.74*100
= 12.52%
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.