The combustion of liquid ethanol (C2H5OH) produces carbon dioxide and water. After 4.62 mL of ethanol (density= 0.789 g/mL) was allowed to burn in the presence of 15.45 g of oxygen gas, 3.71 mL of water (density= 1.00 g/mL) was collected.
a) Find the Limiting Reactant for the reaction (the hint says: Write a balanced equation for the combustion of ethanol.) Is is Ethanol or Oxygen?
b) Determine the theoretical yield of H2O for the reaction.
c) Determine the percent yield of H2O for the reaction.
I understand how to solve theoretical yield equations but this one is really confusing me. Thank you for the help!
a) Find the Limiting Reactant for the reaction (the hint says: Write a balanced equation for the combustion of ethanol.) Is is Ethanol or Oxygen?
b) Determine the theoretical yield of H2O for the reaction.
c) Determine the percent yield of H2O for the reaction.
I understand how to solve theoretical yield equations but this one is really confusing me. Thank you for the help!
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bal. rxn.: C2H5OH + 3 O2 -----> 2CO2 + 3 H2O
moles reactants-
C2H5OH: g=density x vol, g= 4.62ml x 0.789 g/ml= 3.65 g C2H5OH
3.65 g/46 g/mole=0.079 moles ethanol
O2: 15.45 g/16 g/mole=0.966 moles
from bal. rxn., 3 moles of O2 required per mole of C2H5OH
1/3 x .966 moles of O2= 0.322 moles of ethanol required, since there are only 0.079 moles, ethanol is limiting reactant.
from bal. rxn., 1 mole ethanol reacts to form 3 moles water
theo. yield: 0.079 moles x 3 x 18 g/mole H2O= 4.27 g H2O
% yield= actual/theo. x100
% yield=3.71 g/4.27 x100
% yield= 86.9
moles reactants-
C2H5OH: g=density x vol, g= 4.62ml x 0.789 g/ml= 3.65 g C2H5OH
3.65 g/46 g/mole=0.079 moles ethanol
O2: 15.45 g/16 g/mole=0.966 moles
from bal. rxn., 3 moles of O2 required per mole of C2H5OH
1/3 x .966 moles of O2= 0.322 moles of ethanol required, since there are only 0.079 moles, ethanol is limiting reactant.
from bal. rxn., 1 mole ethanol reacts to form 3 moles water
theo. yield: 0.079 moles x 3 x 18 g/mole H2O= 4.27 g H2O
% yield= actual/theo. x100
% yield=3.71 g/4.27 x100
% yield= 86.9
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It seems like these types of questions cause endless trouble for students.
The approach is always the same. It starts with a balanced equation. Then, calculating the number of moles of each to see is one of them is in excess or not.
So a general example:
2A + 3B ----------> C + 4D
Now, if you have two moles of A and three moles of B, you will get a complete reaction, and get one mole of C and 4 moles of D. If you have four moles of B, then B is in excess, but you will still only get one mole of C and 4 moles of D. That is because then the amount of A determines yield.
All right, if you have 2 moles of A, then how many grams is one mole of C and 4 moles of D?
And if one mole of C was 1.0 grams, and you obtained 0.9 grams, then the % yield is 90%.
Hint: since your question has only water and CO2 as products, then either oxygen is the precise amount required, or in excess, since if it was limiting, then CO would be obtained as one of the products.
The approach is always the same. It starts with a balanced equation. Then, calculating the number of moles of each to see is one of them is in excess or not.
So a general example:
2A + 3B ----------> C + 4D
Now, if you have two moles of A and three moles of B, you will get a complete reaction, and get one mole of C and 4 moles of D. If you have four moles of B, then B is in excess, but you will still only get one mole of C and 4 moles of D. That is because then the amount of A determines yield.
All right, if you have 2 moles of A, then how many grams is one mole of C and 4 moles of D?
And if one mole of C was 1.0 grams, and you obtained 0.9 grams, then the % yield is 90%.
Hint: since your question has only water and CO2 as products, then either oxygen is the precise amount required, or in excess, since if it was limiting, then CO would be obtained as one of the products.
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percentage yeild
mass made / maximum amount of product possible x 100
mass made / maximum amount of product possible x 100