Posted by jairo sanchez on Wednesday, March 23, 2011 at 10:59pm.
**i reposted this**
first, make sure the equation is balanced,,
to determine which is limiting, we calculate the amount of product produced by each given:
for 6 mol N2:
6 mol N2 * (2 mol NH3 / 1 mol N2) = 12 mol NH3
for 12 mol H2:
12 mol H2 * (2 mol NH3 / 3 mol H2) = 8 mol NH3
(a) since H2 produced less moles of NH3, H2 is limiting.
(b) thus N2 is in excess.
(c) we get the moles of N2 needed by the given amount of H2:
12 mol H2 * (1 mol N2 / 3 mol H2) = 4 mol N2
6 - 4 = 2 mol N2 left
(d) we already solved this in part (a) ,, the answer we got is 8 mol NH3
(e) percent yield = (actual yield)/(theoretical yield) * 100
percent yield = 6.4 / 8 * 100 = 80%
hope this helps~ :)
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