I need to find the mole ratio of the following and determine which is the limiting reactant and how much of the excess reactant is left over:
.209 mole Acetic Acid
.172 mole Ethyl Alcohol
2007-05-12 13:21:19 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
IMO, both previous answers are mistaken regarding the term "mole ratio." In the first, the mole fraction is given. In the second, the ratio in grams is given (this is a weight ratio, not a mole ratio).
Mole ratio is a term is used in stoichiometric calculations and is obtained via the coefficients of the balanced chemical equation. The mole ratio in this problem is one to one.
one mole of EtOH reacts in proportion with one mole of CH3COOH.
EtOH is the limiting reagent and 0.209 - 0.172 mol of acetic acid remains.
You determine the limiting reagent by dividing each mole amount by the coefficient of that substance from the balanced equation. EtOH (O.172 divided by one) is the answer.
2007-05-12 15:08:15 · answer #1 · answered by ChemTeam 7 · 0⤊ 0⤋
The mole ratio is approximately 60/43 or 30/23.
CH3COOH + C2H5OH ↔ CH3COOC2H5 + H2O
CH3COOH ==> 12*2+1*4+16*2 ≈ 60
C2H5OH ==> 2*12 + 6*1 + 16 ≈ 46
The limiting reactant is CH3COOH with 0.011767 mole (541 mg) C2H5OH left over.
2007-05-12 13:53:56 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋
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2016-05-17 11:08:46 · answer #3 · answered by Anonymous · 0⤊ 0⤋
mole ratio=moles of substance/total moles of compds reacted. X represents mole ratio.
X ace=0.209/0.209+0.172=0.548
X eth=0.172/0.209+0.172=0.451
The limiting reactant is definitely ethyl alcohol. it has fewer moles and it will determine the amount of product to be produced.
Their difference is 0.0370.
0.209-0.172= 0.0370
2007-05-12 13:27:34 · answer #4 · answered by Anonymous · 0⤊ 0⤋