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?
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~ :)
To answer these questions, we need to use stoichiometry and the concept of limiting reactants. Here's how we can find the answers step by step:
a) To determine the limiting reactant, we need to compare the number of moles of N2 and H2 based on the stoichiometric ratio given in the balanced equation. The ratio is 1:3 for N2:H2.
- Number of moles of N2 = 6 mol
- Number of moles of H2 = 12 mol
To find the limiting reactant, we can calculate the amount of moles of NH3 that can be produced from each reactant.
From N2: 6 mol N2 * (2 mol NH3 / 1 mol N2) = 12 mol NH3
From H2: 12 mol H2 * (2 mol NH3 / 3 mol H2) = 8 mol NH3
Since we need 2 mol of NH3 for every 1 mol of N2, and only 8 mol of NH3 can be produced from 12 mol of H2, H2 is the limiting reactant.
b) The chemical in excess is the reactant that is not completely consumed, which in this case is N2.
c) To find the moles of excess reactant left over, we subtract the moles of the limiting reactant used from the initial moles of the excess reactant.
Initial moles of N2 = 6 mol
Initial moles of H2 = 12 mol
Since 12 mol of H2 is used up, we can calculate the moles of N2 remaining:
Remaining moles of N2 = 6 mol - (12 mol * (1 mol N2 / 3 mol H2)) = 6 mol - 4 mol = 2 mol
So, 2 mol of N2 are left over.
d) The moles of NH3 that can be produced are determined by the limiting reactant. Since H2 is the limiting reactant, we can calculate it by multiplying the moles of H2 by the stoichiometric ratio of NH3 to H2:
Moles of NH3 = 12 mol H2 * (2 mol NH3 / 3 mol H2) = 8 mol NH3
Therefore, 8 mol of NH3 can be produced.
e) To calculate the percent yield, we need to compare the actual yield (6.4 mol NH3) to the theoretical yield (8 mol NH3).
Percent yield = (Actual yield / Theoretical yield) * 100
= (6.4 mol / 8 mol) * 100
= 80%
So, the percent yield is 80%.