Question

Calculate the theoretical yield and percent yield of water when the combustion of ethanol produces carbon dioxide and water. After \[4.64{\text{ml}}\] of ethanol (density\[ = 0.789{\text{g/ml}}\]) was allowed to burn in the presence of \[15.75g\] of oxygen gas, \[3.70{\text{ml}}\] of water (density\[ = 1.00{\text{g/ml}}\] was collected.

Answer 1

Hint: For the calculation of theoretical yield and percent yield of water, the first thing is to write a complete balanced equation.
We need to know that the complete balanced equation means the number of atoms on the reactant side is equal to the number of atoms on the product side.

Complete answer:
First step is to write the complete balanced equation for the reaction taking place. The reaction is shown below:
\[{{\text{C}}_2}{{\text{H}}_5}{\text{OH}} + 3{{\text{O}}_2} \to 2{\text{C}}{{\text{O}}_2} + 3{{\text{H}}_2}{\text{O}}\]
From the above reaction we can say that, \[1{\text{ mol}}\]of ethanol reacts with \[3{\text{ mol}}\]of oxygen, to form \[2{\text{ mol}}\]of carbon dioxide and \[3{\text{ mol}}\]of water.
Now, use the formula,
\[{\text{Density}} = \dfrac{{{\text{Mass}}}}{{{\text{Volume}}}}\]
\[{\text{Mass}} = {\text{Density}} \times {\text{volume}}\]
Now, substitute the values, in the above formula
\[{\text{Mass}} = 0.789 \times 4.64 = 3.66g\]
Second step is to calculate the number of moles. Number of moles can be defined as the ratio of given mass to molar mass.
Molar mass of ethanol\[\left( {{{\text{C}}_2}{{\text{H}}_5}{\text{OH}}} \right)\]\[ = 46{\text{g/mol}}\]
Number of moles of ethanol can be calculated as shown below:
\[{\text{n}} = \dfrac{{\text{m}}}{{\text{M}}}\]
\[{\text{n}} = \dfrac{{3.66}}{{46}} = 0.0794{\text{mol}}\]
Number of moles of Oxygen can be calculated as shown below:
Molar mass of oxygen \[\left( {{{\text{O}}_2}} \right) = 32{\text{g/mol}}\]
\[{\text{n}} = \dfrac{{\text{m}}}{{\text{M}}}\]
Now we can substitute the known values we get,
\[{\text{n}} = \dfrac{{3.70}}{{32}} = 0.492{\text{mol}}\]
The number of moles of ethanol is less than that of oxygen, thus the limiting reagent is ethanol.
So, number of moles of water \[ = 0.0794{\text{mol}}\]
Total number of moles of water\[ = 0.0794 \times 3 = 0.238{\text{mol}}\]
\[{\text{mass}} = {\text{n}} \times {\text{M}}\]
Now we can substitute the known values we get,
\[{\text{mass}} = 0.238 \times 18 = 4.29{\text{g}}\]
Percent yield can be calculated by the formula shown below:
Percent yield\[ = \dfrac{{{\text{Actual yield}}}}{{{\text{Theoretical yield}}}} \times 100\]
Percent yield\[ = \dfrac{{3.70}}{{4.29}} \times 100 = 86.3\% \]
Thus, the theoretical yield is \[4.29g\] and percent yield is \[86.3\% \].

Note:
Please keep in mind the less the number of moles, that will act as a limiting reagent. We need to know that the limiting reagent is a reactant which is totally consumed when the chemical reaction is finished. The limiting reagent is used to control the product formation.