if you start with 0.30 mol of Na and and 0.60 mol of N2, and if the actual yield
ID: 967469 • Letter: I
Question
if you start with 0.30 mol of Na and and 0.60 mol of N2, and if the actual yield of Na3N is 6.05g, what is the percent yield? (Hint: balance the equation) Na(s)+N2(g)--> Na3N(s) if you start with 0.30 mol of Na and and 0.60 mol of N2, and if the actual yield of Na3N is 6.05g, what is the percent yield? (Hint: balance the equation) Na(s)+N2(g)--> Na3N(s) if you start with 0.30 mol of Na and and 0.60 mol of N2, and if the actual yield of Na3N is 6.05g, what is the percent yield? (Hint: balance the equation) Na(s)+N2(g)--> Na3N(s)Explanation / Answer
Balanced equation
6Na + N2 --> 2Na3N
From equation, 6 moles of Na and 1 mole of N2 are reacting and generating 2 moles of Na3N.
In question, 0.30 mol of Na and and 0.60 mol of N2 reacting, and the actual yield of Na3N is 6.05g.
0.3 mol of Na will react with 0.05 mol of N2 and generates 0.1 mol of Na3N. In this reaction Na is the limiting reagent. And remaining 0.55 mole of N2 will remain like that only.
Theoretical yield id 0.1 mol of of Na3N.
Convert 0.1 mol of Na3N into grams.
Wt = mole* M.Wt = 0.1*82.97 = 8.297 gm
Theoretical yield is 8.297 gm.
percent yield = (actual yield/ Theoretical yield)*100 = (6.05/8.297)*100 = 72.91%
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