A Balanced equation to represent the reaction between 1-pentanol and ethanoic acid. (Production of an ester)
Its is correct.
It is correct.
Ah, you've put me in a "balanced" situation! Here's the ester-producing reaction between 1-pentanol and ethanoic acid:
1-pentanol + ethanoic acid π€‘ ester + water
C5H12O + C2H4O2 π€‘ C7H14O2 + H2O
Now, wasn't that a truly "balanced" equation?
To balance a chemical equation, we need to ensure that the number of atoms of each element on both sides of the equation is the same. In this reaction, 1-pentanol (C5H11OH) reacts with ethanoic acid (C2H4O2) to produce an ester. The general reaction equation for the formation of an ester is:
Alcohol (R-OH) + Carboxylic Acid (R'-COOH) β Ester (R'-COO-R) + Water (H2O)
Let's balance the equation for the reaction between 1-pentanol and ethanoic acid:
C5H11OH + C2H4O2 β C5H10O2 + H2O
To balance the equation, we start by counting the number of atoms of each element on both sides:
Carbon (C): 5 on the left, 5 on the right
Hydrogen (H): 13 on the left, 12 on the right
Oxygen (O): 3 on the left, 4 on the right
To balance the carbon atoms, we can adjust the coefficient in front of the ester:
C5H11OH + C2H4O2 β 5C5H10O2 + H2O
Now, let's balance the hydrogen and oxygen atoms. There are 11 hydrogen atoms on the left and 14 on the right. To balance the hydrogen, we can adjust the coefficient in front of water:
C5H11OH + C2H4O2 β 5C5H10O2 + 6H2O
Now, we have 12 oxygen atoms on the right, while there are only 3 on the left. To balance the oxygen, we adjust the coefficient in front of ethanoic acid:
C5H11OH + 4C2H4O2 β 5C5H10O2 + 6H2O
Now the equation is balanced, and it represents the reaction between 1-pentanol and ethanoic acid to produce an ester.
CH3CH2CH2CH2CH2-OH + CH3-COOH ==>
CH3CH2CH2CH2CH2OOCCH3 + H2O