# I have a problem that has been driving me crazy trying to solve, and I was wondering if someone could help. The problem is this: "Given the reaction: CuSO4 + 4 NH3 ----> Cu(NH3)4SO4, if 10 grams of CuSO4 reacts with 30 grams of NH3, what is the theoretical yield of Cu(NH3)4SO4, what is the limiting reactant, how many grams of the excess reactant is left over, and if the actual yield is 12.6 grams, what is the percent yield?" I don't know if it's hard or not, but I keep getting stuck after finding the theoretical yield (176.24?). Could someone help?

## Post what you have done. Getting a theoretical yield of 176 grams out of 40 grams of reactants is visionary.

## I add up the atomic masses and get the theoretical yield, it's finding the limiting reactant that is getting to me.

## What I mean to say is I add up the atomic masses of CuSO4 + 4 NH3 and get the theoretical yield of 176.24. Or am I finding it wrong?

## Let me help you on this...

To find the limiting reactant first of all.

-You have to first find the moles you have of each reactant.

- then go and find out how much of each reactant do you need to consume the other reactant

- then compare it to how much moles you do have of each and the one that needs more to consume than you have is the limiting reactant

I'll start with this and PLEASE post your work as I'm NOT going to do all the work =) (I'll check later)

## Of course, I'd be happy to help you with that problem!

To solve this problem, we'll need to follow a step-by-step approach. Let's break it down:

Step 1: Calculate the molar masses

First, we need to calculate the molar masses of each compound involved in the reaction.

The molar mass of CuSO4 (copper(II) sulfate) can be calculated as follows:

- Cu: 63.55 g/mol

- S: 32.07 g/mol

- O (4 atoms): 16.00 g/mol

Adding these values together, the molar mass of CuSO4 is 159.61 g/mol.

Next, let's calculate the molar mass of NH3 (ammonia):

- N: 14.01 g/mol

- H (3 atoms): 1.01 g/mol

Adding these values together, the molar mass of NH3 is 17.03 g/mol.

Step 2: Convert grams to moles

Now, let's convert the given masses of CuSO4 and NH3 into moles.

Using the formula: moles = mass (g) / molar mass (g/mol)

For CuSO4: moles = 10 g / 159.61 g/mol = 0.0627 mol (rounded to four decimal places)

For NH3: moles = 30 g / 17.03 g/mol = 1.762 mol (rounded to three decimal places)

Step 3: Determine the limiting reactant

To identify the limiting reactant, we compare the mole ratios between CuSO4 and NH3 in the balanced equation:

CuSO4 + 4 NH3 ---> Cu(NH3)4SO4

From the equation, we can see that one mole of CuSO4 reacts with four moles of NH3.

Comparing the moles obtained in step 2, we have:

- CuSO4: 0.0627 mol

- NH3: 1.762 mol

To find the limiting reactant, we divide each number of moles by the stoichiometric coefficient in the balanced equation:

For CuSO4: 0.0627 mol / 1 = 0.0627 mol

For NH3: 1.762 mol / 4 = 0.4405 mol

We see that CuSO4 has the smaller value, indicating that it is the limiting reactant.

Step 4: Determine the theoretical yield

To find the theoretical yield of Cu(NH3)4SO4, we need to use the stoichiometry of the balanced equation and the limiting reactant.

From the balanced equation, we know that one mole of CuSO4 reacts to form one mole of Cu(NH3)4SO4.

Therefore, the theoretical yield of Cu(NH3)4SO4 is equal to the number of moles of CuSO4, which is 0.0627 mol, multiplied by its molar mass:

Theoretical yield = 0.0627 mol * (molar mass of Cu(NH3)4SO4)

Substituting the molar masses of the elements in the compound:

- Cu: 63.55 g/mol

- N: 14.01 g/mol

- H (12 atoms): 1.01 g/mol

- S: 32.07 g/mol

- O (4 atoms): 16.00 g/mol

We can calculate the molar mass of Cu(NH3)4SO4 as 275.79 g/mol.

Therefore, the theoretical yield is 0.0627 mol * 275.79 g/mol = 17.27 grams (rounded to two decimal places).

Step 5: Determine the excess reactant and calculate the percent yield

To determine the excess reactant, we need to compare the moles of the limiting reactant (0.0627 mol) with the moles of the other reactant initially present (NH3 in this case).

NH3 moles initially present = 1.762 mol

NH3 moles consumed = 0.0627 mol (which reacted with the limiting reactant)

NH3 moles remaining = NH3 moles initially present - NH3 moles consumed = 1.762 mol - 0.0627 mol = 1.6993 mol (rounded to four decimal places)

To find the mass of NH3 remaining, we multiply the remaining moles by its molar mass:

mass remaining = 1.6993 mol * 17.03 g/mol = 28.95 grams (rounded to two decimal places)

Finally, to calculate the percent yield, we use the formula:

percent yield = (actual yield / theoretical yield) * 100

For this problem, the actual yield is given as 12.6 grams.

percent yield = (12.6 g / 17.27 g) * 100 = 73.0% (rounded to one decimal place)

So, to summarize:

- The theoretical yield of Cu(NH3)4SO4 is 17.27 grams.

- The limiting reactant is CuSO4.

- The excess reactant is NH3, with 28.95 grams remaining.

- The percent yield is 73.0%.