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?

To determine the limiting reactant, we need to compare the moles of N2 and H2 and see which one is used up completely.

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

a) To find the limiting reactant, we need to calculate the number of moles of NH3 that can be produced from each reactant:

From the balanced equation: N2 + 3H2 -> 2NH3

For N2: 1 mole of N2 produces 2 moles of NH3.
So, 6 moles of N2 will produce (6 mol N2) x (2 mol NH3 / 1 mol N2) = 12 moles of NH3.

For H2: 3 moles of H2 produce 2 moles of NH3.
So, 12 moles of H2 will produce (12 mol H2) x (2 mol NH3 / 3 mol H2) = 8 moles of NH3.

We can see that the number of moles of NH3 that can be produced is limited by the H2 reactant, as H2 can only produce 8 moles of NH3, while N2 can produce 12 moles.

Therefore, b) H2 is the limiting reactant.

c) To find the number of moles of excess reactant left over, we need to subtract the moles of the limiting reactant used from the initial moles.

The H2 is the limiting reactant, so all of it will be used up in the reaction. Therefore, no excess H2 is left over.

d) From the balanced equation, we can see that 3 moles of H2 are required to produce 2 moles of NH3.

Since we have 12 moles of H2, we can calculate the moles of NH3 produced as follows:

12 moles H2 * (2 moles NH3 / 3 moles H2) = 8 moles NH3

So, d) 8 moles of NH3 can be produced.

e) If only 6.4 moles of NH3 are produced, we can calculate the percent yield by dividing the actual yield (6.4 moles) by the theoretical yield (calculated in part d) of 8 moles) and multiplying by 100:

Percent yield = (Actual yield / Theoretical yield) x 100
Percent yield = (6.4 moles / 8 moles) x 100 = 80%

Therefore, e) the percent yield is 80%.

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

a) First, let's calculate the number of moles of NH3 that can be produced from each reactant.

For N2:
Given: 6 mol N2
From the balanced equation, the stoichiometric ratio between N2 and NH3 is 1:2. Therefore, for every 1 mole of N2, 2 moles of NH3 are produced.
So, the moles of NH3 that can be produced from N2 = 6 mol N2 * (2 mol NH3 / 1 mol N2) = 12 mol NH3

For H2:
Given: 12 mol H2
From the balanced equation, the stoichiometric ratio between H2 and NH3 is 3:2. Therefore, for every 3 moles of H2, 2 moles of NH3 are produced.
So, the moles of NH3 that can be produced from H2 = 12 mol H2 * (2 mol NH3 / 3 mol H2) = 8 mol NH3

The limiting reactant is the one that produces the lesser amount of product. In this case, H2 produces 8 mol NH3, while N2 produces 12 mol NH3. Therefore, H2 is the limiting reactant.

b) The chemical that is in excess is the one that is not the limiting reactant. In this case, N2 is in excess.

c) To calculate the moles of excess reactant left over, subtract the moles used from the initial moles of the excess reactant. From the given information, we know that 6 mol of H2 were used (limiting reactant), so the initial moles of H2 are still 12 mol. Therefore, the moles of excess reactant left over (N2) is 6 mol (12 mol - 6 mol = 6 mol).

d) The moles of NH3 that can be produced is determined by the limiting reactant. Since H2 is the limiting reactant, we can produce 8 mol of NH3.

e) To calculate the percent yield, we divide the actual yield (given as 6.4 mol NH3) by the theoretical yield (calculated as 8 mol NH3) and multiply by 100%.
Percent yield = (actual yield / theoretical yield) * 100% = (6.4 mol / 8 mol) * 100% ≈ 80%

Therefore:
a) The limiting reactant is H2.
b) The chemical in excess is N2.
c) The moles of excess reactant left over is 6 mol.
d) The moles of NH3 that can be produced is 8 mol.
e) The percent yield is approximately 80%.