Question

6mols of N2 are mixed with 12mol H2, according to the following equation.

N2+3H2----> 2NH3

a) which is the limiting reactant?

B) which chemical is in excess?

c) how many moles of excess reactant is left over?

d) how many moles of NH3 can be produced?

e)if only 6.4 mols of NH3 are produced What is the percent yield?

Answer 1

**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~ :)