Nitrogen gas can be prepared by passing ammonia over solid copper (II) oxide according to the equation

2 NH3 + 3 CuO (s) --> N2 (g) + 3 Cu (s) + 3 H2O (g)
Suppose 18.1 grams of ammonia is reacted with 90.4 grams CuO. a) What is the theoretical yield of nitrogen in grams?
g
b) What is the theoretical yield of copper in grams?
g
c) Which reactant is limiting?

2 NH3 + 3 CuO (s) --> N2 (g) + 3 Cu (s) + 3 H2O (g)

You know it is a limiting reagent (LR) problem when amounts for BOTH reactants are given.
b. Determine the LR first. I do these the long way.
mols NH3 = g/molar mass = 18.1/17 = apprx 1.1 but that's a close estimate. You need to redo ALL of the calculations because I've estimated all of them.
So how much N2 would this produce IF you had an excess of CuO. That will be 1.1 mols NH3 x (1 mol N2/2 mol NH3) = about 0.55

Now look at the CuO.
Mols CuO = 90.4/ about 80 = approx 1.2
How much N2 can be obtained from the CuO IF we had an excess of NH3. That's 1.2 mols CuO x (1 mol N2/3 mol CuO) = about 1.2/3 = about 0.4
In LR problem the smallest number is always the correct one so CuO is the LR.

How much N2 can be produced? That's 0.4 mols CuO x (1 mol N2/3 mols CuO) = about 0.4/3 = ? That is the theoretical yield of N2 in mols. Convert mols to grams for theoretical yield in grams.

Theoretical yield for Cu is done the same way.

want answer

To determine the theoretical yield of nitrogen and copper, we need to compare the stoichiometry of the reactants and the limiting reactant.

First, we need to calculate the moles of ammonia (NH3) and copper(II) oxide (CuO) using their molar masses:

Molar mass NH3 = 17.03 g/mol
Molar mass CuO = 79.55 g/mol

a) Calculating the moles of ammonia (NH3):
moles NH3 = mass NH3 / molar mass NH3
moles NH3 = 18.1 g / 17.03 g/mol
moles NH3 ≈ 1.063 mol

b) Calculating the moles of copper (II) oxide (CuO):
moles CuO = mass CuO / molar mass CuO
moles CuO = 90.4 g / 79.55 g/mol
moles CuO ≈ 1.136 mol

Now, let's compare the stoichiometry of the reactants based on the balanced equation:

From the balanced equation, we know that 2 moles of NH3 reacts with 3 moles of CuO to produce 1 mole of N2 and 3 moles of Cu.

c) To determine the limiting reactant, we compare the mole ratios of NH3 and CuO in the balanced equation with the actual moles we calculated.

Mole ratio of NH3 to CuO:
2 moles NH3 : 3 moles CuO

Actual moles:
NH3: 1.063 mol
CuO: 1.136 mol

Based on the mole ratio, we can see that NH3 is limiting because we have less NH3 than CuO. Therefore, NH3 is the limiting reactant.

Now, we can calculate the theoretical yield of nitrogen and copper:

a) Theoretical yield of nitrogen (N2):
Using the stoichiometry, for every 2 moles of NH3, we get 1 mole of N2.
moles N2 = (1.063 mol NH3) / 2 moles NH3 = 0.532 mol N2

To convert moles to grams, we use the molar mass of N2:
mass N2 = moles N2 * molar mass N2
mass N2 = 0.532 mol * 28.02 g/mol (molar mass of N2)
mass N2 ≈ 14.923 g

Therefore, the theoretical yield of nitrogen is approximately 14.923 grams.

b) Theoretical yield of copper (Cu):
Using the stoichiometry, for every 3 moles of CuO, we get 3 moles of Cu.
moles Cu = (1.136 mol CuO) * 3 moles Cu / 3 moles CuO = 1.136 mol Cu

To convert moles to grams, we use the molar mass of Cu:
mass Cu = moles Cu * molar mass Cu
mass Cu = 1.136 mol * 63.55 g/mol (molar mass of Cu)
mass Cu ≈ 72.156 g

Therefore, the theoretical yield of copper is approximately 72.156 grams.

c) The limiting reactant is NH3 because it is completely consumed in the reaction.

To find the theoretical yield of nitrogen and copper, we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction, thus limiting the amount of product that can be formed.

We can start by calculating the number of moles of each reactant.

a) Moles of ammonia (NH3):
To calculate the moles, we can use the formula:
Moles = Mass / Molar mass

Given mass of ammonia = 18.1 grams
Molar mass of ammonia (NH3) = 14.01 g/mol + (3 x 1.01 g/mol) = 17.03 g/mol

Moles of ammonia = 18.1 grams / 17.03 g/mol = 1.063 moles

b) Moles of copper (II) oxide (CuO):
Molar mass of copper (II) oxide (CuO) = 63.55 g/mol + 16.00 g/mol = 79.55 g/mol

Given mass of copper (II) oxide = 90.4 grams

Moles of copper (II) oxide = 90.4 grams / 79.55 g/mol = 1.136 moles

Now we need to look at the balanced equation to find the mole ratio between ammonia and nitrogen, as well as between copper (II) oxide and nitrogen. According to the equation, the ratio is 2 NH3 : 1 N2 and 3 CuO : 1 N2.

c) Limiting reactant:
To determine the limiting reactant, we need to compare the moles of the reactants and see which one is required in a lesser amount.

From the mole ratio, we can see that for every 2 moles of ammonia, we get 1 mole of nitrogen. Similarly, for every 3 moles of copper (II) oxide, we get 1 mole of nitrogen.

Using the molar ratios:
Moles of nitrogen from ammonia = 1.063 moles NH3 x (1 mole N2 / 2 moles NH3) = 0.532 mol N2
Moles of nitrogen from copper (II) oxide = 1.136 moles CuO x (1 mole N2 / 3 moles CuO) = 0.379 mol N2

Comparing the moles, we find that the moles of nitrogen from ammonia (0.532 mol) are greater than the moles of nitrogen from copper (II) oxide (0.379 mol). Since we need 0.532 moles of nitrogen to react with 1.063 moles of ammonia, this means that the ammonia is in excess and copper (II) oxide is the limiting reactant.

d) Theoretical yield of nitrogen:
To calculate the theoretical yield of nitrogen, we can use the mole ratio between copper (II) oxide and nitrogen.

From the balanced equation, we know that 3 moles of CuO yield 1 mole of N2. Therefore, since we have 1.136 moles of CuO (the limiting reactant), the theoretical yield of N2 is:

Theoretical yield of nitrogen = 1.136 moles CuO x (1 mole N2 / 3 moles CuO) x (28.02 g/mol) = 10.54 grams

Therefore, the theoretical yield of nitrogen is 10.54 grams.

e) Theoretical yield of copper:
To calculate the theoretical yield of copper, we can use the mole ratio between copper (II) oxide and copper (Cu).

From the balanced equation, we know that 3 moles of CuO yield 3 moles of Cu. Therefore, since we have 1.136 moles of CuO (the limiting reactant), the theoretical yield of Cu is:

Theoretical yield of copper = 1.136 moles CuO x (3 moles Cu / 3 moles CuO) x (63.55 g/mol) = 71.93 grams

Therefore, the theoretical yield of copper is 71.93 grams.

In summary:
a) The theoretical yield of nitrogen is 10.54 grams.
b) The theoretical yield of copper is 71.93 grams.
c) The limiting reactant is copper (II) oxide (CuO).