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?

**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~ :)

a) Well, if we want to figure out the limiting reactant, we need to see which one it is that we're going to run out of first. So, let's take a look at the balanced equation and compare the ratio of N2 to H2. It's 1:3, right? Since we have 6 moles of N2 and 12 moles of H2, it looks like we have enough H2 for both reactants. But, if we divide 6 by 1, we get 6, and if we divide 12 by 3, we get 4. So, it looks like our limiting reactant is N2. Poor N2, it just can't keep up with the hydrogen!

b) Now that we know N2 is the limiting reactant, that means H2 is in excess. It's like H2 is showing off, saying "Hey, N2, look at me with all my extra moles!"

c) Since we know H2 is in excess, that means we'll have some leftover after the reaction is complete. To find out how much, we'll have to do a little math. We started with 12 moles of H2, and since the ratio of N2 to H2 is 1:3, we know that for every 1 mole of N2 used, we need 3 moles of H2. So, if we use up all 6 moles of N2, that means we would have used 2 moles of H2 (since the ratio is 1 N2 to 3 H2). That leaves us with 12 - 2 = 10 moles of H2 leftover. Looks like we have a little H2 party going on!

d) Alright, time to figure out how many moles of NH3 can be produced. Since the balanced equation tells us that for every 1 mole of N2, we get 2 moles of NH3, and we know we have 6 moles of N2 (our limiting reactant), that means we can produce 2 * 6 = 12 moles of NH3.

e) Ah, the tricky one. Percent yield is all about comparing the actual yield to the theoretical yield. Theoretical yield is what we calculated earlier, which is 12 moles of NH3. Actual yield is what was given, which is 6.4 moles. To calculate percent yield, we use the formula (actual yield / theoretical yield) * 100. So, (6.4 / 12) * 100 = 53.3%. Not too shabby, but it seems like the reaction could've been a bit more productive!

To determine the limiting reactant, excess reactant, moles of excess reactant left over, moles of NH3 produced, and the percent yield, we need to use the stoichiometry of the balanced equation.

Given:
Moles of N2 = 6 mol
Moles of H2 = 12 mol

a) To find the limiting reactant, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation. The stoichiometric ratio between N2 and NH3 is 1:2, and between H2 and NH3 is 3:2.

For N2: The moles of NH3 that can be formed from 6 mol of N2 = 6 mol × (2 mol NH3/1 mol N2) = 12 mol NH3
For H2: The moles of NH3 that can be formed from 12 mol of H2 = 12 mol × (2 mol NH3/3 mol H2) = 8 mol NH3

We see that H2 is the limiting reactant since it produces fewer moles of NH3 compared to N2.

b) The chemical in excess is N2 since we had more moles of N2 (6 mol) than required compared to the stoichiometric ratio.

c) To determine how many moles of excess reactant is left over, we need to find how much H2 was not consumed in the reaction.
The moles of H2 that reacted = 12 mol (amount used to form NH3 from the limiting reactant)
The moles of excess H2 left over = 12 mol - moles of H2 that reacted = 12 mol - 8 mol = 4 mol.

d) To find how many moles of NH3 can be produced, we use stoichiometry:
The moles of NH3 produced = moles of the limiting reactant × (2 mol NH3/3 mol H2) = 8 mol × (2 mol NH3/3 mol H2) = 5.33 mol

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
Percent yield = (6.4 mol / 8 mol) × 100 = 80%

To answer these questions, we need to use stoichiometry and compare the given moles with the ratios in the balanced chemical equation.

a) To determine the limiting reactant, we compare the moles of N2 and H2 and calculate the moles of NH3 that can be produced from each reactant.

For N2:
Moles of NH3 = (6 mol N2) * (2 mol NH3 / 1 mol N2) = 12 mol NH3

For H2:
Moles of NH3 = (12 mol H2) * (2 mol NH3 / 3 mol H2) = 8 mol NH3

Hence, the limiting reactant is H2 because it produces fewer moles of NH3.

b) To determine which chemical is in excess, we compare the moles of the excess reactant with the moles required from the limiting reactant.

For H2:
Excess moles of H2 = (12 mol H2) - (8 mol NH3 * (3 mol H2 / 2 mol NH3)) = 12 - 4.5 = 7.5 mol H2

Therefore, H2 is in excess.

c) To calculate the moles of excess reactant left over, we subtract the moles of the reactant used from the total moles of the excess reactant:

Moles of excess reactant left over = (12 mol H2) - (8 mol NH3 * (3 mol H2 / 2 mol NH3)) = 12 - 4.5 = 7.5 mol H2

d) The balanced chemical equation tells us that 1 mole of N2 reacts to produce 2 moles of NH3. Therefore, the number of moles of NH3 that can be produced is equal to the moles of the limiting reactant (H2 in this case):

Moles of NH3 produced = 8 mol NH3

e) To calculate the percent yield, we compare the actual yield (6.4 mol NH3) with the theoretical yield (8 mol NH3) and calculate the percentage:

Percent yield = (Actual yield / Theoretical yield) * 100
Percent yield = (6.4 mol NH3 / 8 mol NH3) * 100 = 80%