<u>Answer:</u>
<u>For a:</u> The limiting reactant is ethanol.
<u>For b:</u> The theoretical yield of water is 4.27 grams.
<u>For c:</u> The percentage yield of water is 86.6 %
<u>Explanation:</u>
<u>For ethanol:</u>
To calculate the mass of ethanol, we use the equation:
......(1)
Density of ethanol = 0.789 g/mL
Volume of ethanol = 4.60 mL
Putting values in equation 1, we get:
To calculate the number of moles, we use the equation:
.....(2)
Given mass of ethanol = 3.63 g
Molar mass of ethanol = 46.1 g/mol
Putting values in equation 2, we get:
Given mass of oxygen gas = 15.50 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 2, we get:
The chemical equation for the combustion of ethanol follows:
By Stoichiometry of the reaction:
1 mole of ethanol reacts with 3 moles of oxygen gas
So, 0.079 moles of ethanol will react with = of oxygen gas
As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.
Thus, ethanol is considered as a limiting reagent because it limits the formation of product.
Hence, the limiting reactant is ethanol.
By Stoichiometry of the reaction:
1 mole of ethanol produces 3 moles of water
So, 0.079 moles of ethanol will produce = of water
Now, calculating the theoretical yield of water by using equation 1:
Molar mass of water = 18 g/mol
Moles of water = 0.237 moles
Putting values in equation 2, we get:
Hence, the theoretical yield of water is 4.27 grams.
Calculating the mass of water, by using equation 1, we get:
Density of water = 1.00 g/mL
Volume of water = 3.70 mL
Putting values in equation 1, we get:
To calculate the percentage yield of water, we use the equation:
Experimental yield of water = 3.70 g
Theoretical yield of water = 4.27 g
Putting values in above equation, we get:
Hence, the percentage yield of water is 86.6 %