%yield and limiting reagents (pls show work on all probelms)

Question

2. Fe 3 O 4 + 4 H 2 → 3Fe + 4H 2 O

a. If you start with 0.122 g of H 2 , how much H 2 O do you theoretically

make?

b. You make 0.745 g of H 2 O in the lab, what is your percent yield?

3. 2Al+ 3Cl 2 → 2AlCl 3

How many grams of AlCl 3 could be produced from 34.0 g Al and 39.0 g

Cl 2 ?

4. N 2 + 3H 2 → 2NH 3

a. How many grams of NH 3 will be produced if you have 55.2 g N 2 and

89.7 g of H 2 ?

b. What are your limiting reagents?

5. C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O

a. If you have 17.6 g of C 3 H 8 and 36.2 g of O 2 , how much CO 2 will be

made?

b. What is your limiting reagent?

Answer 1

I won't do all for you, but will do the first problem and will do one with limiting reagents to show you the process. Hopefully, you'll be able to do the others.

2. Fe3O4 + 4H2 → 3Fe + 4H2O

a. If you start with 0.122 g of H2 , how much H2O do you theoretically

make? 0.122 g H2 x 1 mol H2/2 g x 4 mol H2O / 4 mol H2 x 18 g H2O/mol H2O = 1.09g H2O theoretical yield

b. You make 0.745 g of H2O in the lab, what is your percent yield? actual yield/theoretical yield (x100) gives percent yield: 0.745 g / 1.09 g (x100) = 68.3% yield

An easy way to find the limiting reagent is to divide the moles of each reagent by its coefficient in the balanced equation and whichever comes out less is the limiting reagent. Here's an example using question #5 above.

5.) C3H8 + 5O2 → 3CO2 + 4H2O

17.6 g of C3H8 and 36.2 g of O2

For C3H8: 17.6 g x 1 mol/44 g = 0.4 moles (÷ 1 = 0.4)

For O2: 36.2 g x 1 mol/32 g = 1.13 moles (÷ 5 = 0.23)

So 0.23 is less than 0.4 and O2 is the limiting reagent.

Then use 1.13 moles O2 to find the amount of product that can be formed.

Answer 2

Hello, Gabrielle,

a. If you start with 0.122 g of H 2 , how much H 2 O do you theoretically

make? Calculate the moles of H_2, which would be (0.122g/2g/mole), or 0.061 moles of H₂. The balanced equation says we'll get 4 moles of water for every four moles of H₂ (1 to 1), so we should get 0.061 moles of H₂O. Convert that to grams by multiplying by water's molar mass of 18 g/moles.

b. You make 0.745 g of H 2 O in the lab, what is your percent yield? Take the answer from (1.) and divide that by 0.745 g to get percent yield. I judge around 70%, just looking at the numbers.

3. 2Al+ 3Cl 2 → 2AlCl 3

How many grams of AlCl 3 could be produced from 34.0 g Al and 39.0 g

Cl 2 ? Use the same process as above. Calculate the moles of Al in 34.0 grams. Since it is in a 1:1 relationship with the moles of AlCl₃, that will be the number of moles of the product produced. Then convert that into grams by multiplying by the molar mass of AlCl_3.

4. N 2 + 3H 2 → 2NH 3

a. How many grams of NH 3 will be produced if you have 55.2 g N 2 and

89.7 g of H 2 ? Use the same process. But now we have a molar ratio of 2:1 between the NH₃ and N₂, so multiply the value of moles you get for the N₂ by 2 to get moles of NH₃. Before you do that, however, we need to find out if there is enough H₂ for this reaction. The equation says we need three moles of H₂ for every one mole of N₂. Calculate how many moles we have of both the H₂ and N2 by dividing their masses by their molar mass. I get 1.97 moles for N₂ and 44.85 moles of H₂. We need three times as much hydrogen than nitrogen, and we can see we have far more than that. That makes nitrogen the limiting reagent. Use the moles of N₂ in calculating the grams of NH₃ using the technique above.

b. What are your limiting reagents? Answered just above.

5. C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O

a. If you have 17.6 g of C 3 H 8 and 36.2 g of O 2 , how much CO 2 will be

made? I've run out of time. Sorry. Use the same principles as the previous problem. Convert to moles and determine which reagent is limiting before using that reagent to calculate the amount of CO₂ made.

b. What is your limiting reagent?

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I hope this helps,

Bob