When propyl alcohol is treated with acid, the initially formed intermediate is known as an oxonium ion. There is a scheme of a r
1 answer:
Answer:
See explanation and image attached
Explanation:
The question has to do with the protonation of an alcohol. We know that oxygen has two lone pairs of electrons. Now a lone pair of electrons can pick up a proton leading to the formation of a coordinate bond between the proton and the oxygen atom.
The oxygen atom in the alcohol now has a positive charge as shown in the image attached to this answer. All the bonds remain single.
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The question is incomplete, here is the complete question:
Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water. If 12.5 g of water is produced from the reaction of 72.6 g of sulfuric acid and 77.0 g of sodium hydroxide, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.
<u>Answer:</u> The percent yield of water in the reaction is 46.85 %.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For NaOH:</u>
Given mass of NaOH = 77.0 g
Molar mass of NaOH = 40 g/mol
Putting values in equation 1, we get:
- <u>For sulfuric acid:</u>
Given mass of sulfuric acid = 72.6 g
Molar mass of sulfuric acid = 98 g/mol
Putting values in equation 1, we get:
The chemical equation for the reaction of NaOH and sulfuric acid follows:
By Stoichiometry of the reaction:
1 mole of sulfuric acid reacts with 2 moles of NaOH
So, 0.741 moles of sulfuric acid will react with = of NaOH
As, given amount of NaOH is more than the required amount. So, it is considered as an excess reagent.
Thus, sulfuric acid is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of sulfuric acid produces 2 moles of water
So, 0.741 moles of sulfuric acid will produce = of water
Now, calculating the mass of water from equation 1, we get:
Molar mass of water = 18 g/mol
Moles of water = 1.482 moles
Putting values in equation 1, we get:
To calculate the percentage yield of water, we use the equation:
Experimental yield of water = 12.5 g
Theoretical yield of water = 26.68 g
Putting values in above equation, we get:
Hence, the percent yield of water in the reaction is 46.85 %.