# Need some chemistry help on calculating theoretical yields...

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#1
Hello everyone.

I have an assignment to do for my adult learner course, I am having some difficulty with stoichiometry and calculating theoretical yields and atom economies.

I have two course work questions. The first is this:

Calculate the percentage yield and atom economy for the following reaction:

2al + 3cl2 = 2alcl3
12.8g of aluminium reacted with excess chlorine and gave 30.2g of aluminium chloride.

I have worked out how much of a mole I have of aluminium.

12.8/27= 0.47 of a mole of al

And how much of a mole of aluminium chloride I have.

30.2/133.5= 0.23 of a mole of alcl3

0.23 divided by 0.47 times 100= 48.9% theoretical yield? Am I working this out correctly?

Thanks so much for any help

Ollie
0
1 year ago
#2
yes correct
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#3
(Original post by BobbJo)
yes correct
Thank you.

And the atom economy would be 100% as the Alcl3 is the only product produced?

I have another similar question which gets a big more taxing as well I will post it in a moment.
0
1 year ago
#4
(Original post by modifiedgenes)
Thank you.

And the atom economy would be 100% as the Alcl3 is the only product produced?

I have another similar question which gets a big more taxing as well I will post it in a moment.
yes correct again
0
#5
Cuo + h2 = Cu + H2o

20g of copper oxide with excess hydrogen gave 11.3g of copper.

20g divided by 79.55 (16 + 63.55) = 0.25 moles

11.3g divided by 63.5 = 0.18 moles

0.18 divided by 0.25= 0.72 or 72% yield

Atom economy

63.5 divided by 81.5 (63.5 + 2 hydrogen and 1 oxygen) = 77.9% atom economy.
0
#6
Thank you so much for your help.
0
1 year ago
#7
(Original post by modifiedgenes)
Cuo + h2 = Cu + H2o

20g of copper oxide with excess hydrogen gave 11.3g of copper.

20g divided by 79.55 (16 + 63.55) = 0.25 moles

11.3g divided by 63.5 = 0.18 moles

0.18 divided by 0.25= 0.72 or 72% yield

Atom economy

63.5 divided by 81.5 (63.5 + 2 hydrogen and 1 oxygen) = 77.9% atom economy.
you round excessively. the first question the answer is 47.7% for the % yield.

for this one it is 70.7%.

keep intermediate values to more decimal places

atom economy is correct
1
#8
(Original post by BobbJo)
you round excessively. the first question the answer is 47.7% for the % yield.

for this one it is 70.7%.

keep intermediate values to more decimal places

atom economy is correct
I have another chemistry related question if you would not mind taking a look, I just need to check my answers are correct.

25.0g of ethanol C2H5OH was oxidised, 29.5g of ethanoic acid CH3COOH were obtained

formula mass of ethanol= 46g
formula mass ethanoic acid= 60

moles of ethanol used
25/46= 0.54347 moles

moles of ethanoic acid produced
29.9/60= 0.49166666667 moles

0.4916666666/0.54347 x 100= 90.46% percentage yield.

Thanks again.
0
1 year ago
#9
(Original post by modifiedgenes)
I have another chemistry related question if you would not mind taking a look, I just need to check my answers are correct.

25.0g of ethanol C2H5OH was oxidised, 29.5g of ethanoic acid CH3COOH were obtained

formula mass of ethanol= 46g
formula mass ethanoic acid= 60

moles of ethanol used
25/46= 0.54347 moles

moles of ethanoic acid produced
29.9/60= 0.49166666667 moles

0.4916666666/0.54347 x 100= 90.46% percentage yield.

Thanks again.
Correct! [90.46666 could be rounded to 3 sf as the data are to 3 sf]
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#10
(Original post by BobbJo)
Correct! [90.46666 could be rounded to 3 sf as the data are to 3 sf]
Thank you, I have rounded it to 4 sf as it is a percentage.

I am still not certain of this question, 100g of calcium carbonate gave 45g of calcium oxide.

Formula mass of CaCO3= 100.1g
Formula mass of CaO= 56.1g
Formula mass of CO2= 44g

Moles of CaCO3 used= 0.999
Moles of CaO produced= 0.8021

0.8021/0.999 x 100= 80.92% yield

atom economy

56.1 + 44= 100.1

56.1/100.1 x 100= 56% atom economy.

Thanks again.
0
1 year ago
#11
(Original post by modifiedgenes)
Thank you, I have rounded it to 4 sf as it is a percentage.

I am still not certain of this question, 100g of calcium carbonate gave 45g of calcium oxide.

Formula mass of CaCO3= 100.1g
Formula mass of CaO= 56.1g
Formula mass of CO2= 44g

Moles of CaCO3 used= 0.999
Moles of CaO produced= 0.8021

0.8021/0.999 x 100= 80.92% yield

atom economy

56.1 + 44= 100.1

56.1/100.1 x 100= 56% atom economy.

Thanks again.
surely you mean 80.29%

atom economy is correct
0
#12
(Original post by BobbJo)
surely you mean 80.29%

atom economy is correct
Thank you so much. I have an absolutely stinking cold and I am struggling to finish this assignment though it isn't due for over a week. Really finding it difficult to sleep at night as a result and I really need to get these calculations right.

Thank you again.
0
1 year ago
#13
(Original post by modifiedgenes)
Thank you so much. I have an absolutely stinking cold and I am struggling to finish this assignment though it isn't due for over a week. Really finding it difficult to sleep at night as a result and I really need to get these calculations right.

Thank you again.
Good luck to you! Hard work always pays off 0
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