Question

The combustion of liquid ethanol (C2H5OH) produces carbon dioxide and water. After 4.60 mL of ethanol (density=0.789g/ml) was allowed to burn in the presence of 15.50 g of oxygen gas, 3.70 mL of water (density=1.00g/ml) was collected.

a) Determine the limiting reactant for the reaction. (Hint: Write a balanced equation for the combustion of ethanol.)
b) Determine the theoretical yield of H2O for the reaction.
c) Determine the percent yield of H2O for the reaction.

Answer 1

Answer:

a. Ethanol.

b. 4.21 mL H2O.

c. 87.8%

Explanation:

Hello,

- For the limiting reactant, the available amount of ethanol is less than the ideal amount of ethanol that would react with the given amount of oxygen, that's why it is the limiting one.

- You will find the calculations in the attached picture; take into account that the molecular mass of water is 18 g/mol and ethanol 46 g/mol.

- The percent yield is allowed to be computed in terms of volume as well, because the relationships are conserved.

Best regards!

Answer 2

Answer:

For a: The limiting reactant is ethanol.

For b: The theoretical yield of water is 4.27 grams.

For c: The percentage yield of water is 86.6 %

Explanation:

• For a:

For ethanol:

To calculate the mass of ethanol, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$     ......(1)

Density of ethanol = 0.789 g/mL

Volume of ethanol = 4.60 mL

Putting values in equation 1, we get:

$0.789g/mL=\frac{\text{Mass of ethanol}}{4.60mL}\\\\\text{Mass of ethanol}=(0.789g/mL\times 4.60mL)=3.63g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(2)

Given mass of ethanol = 3.63 g

Molar mass of ethanol = 46.1 g/mol

Putting values in equation 2, we get:

$\text{Moles of ethanol}=\frac{3.63g}{46.1g/mol}=0.079mol$

• For oxygen gas:

Given mass of oxygen gas = 15.50 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 2, we get:

$\text{Moles of oxygen gas}=\frac{15.50g}{32g/mol}=0.484mol$

The chemical equation for the combustion of ethanol follows:

$C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O$

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 = $\frac{3}{1}\times 0.079=0.237mol$ 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.

• For b:

By Stoichiometry of the reaction:

1 mole of ethanol produces 3 moles of water

So, 0.079 moles of ethanol will produce = $\frac{3}{1}\times 0.079=0.237mol$ 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:

$0.237mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.237mol\times 18g/mol)=4.27g$

Hence, the theoretical yield of water is 4.27 grams.

• For c:

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:

$1.00g/mL=\frac{\text{Mass of water}}{3.70mL}\\\\\text{Mass of water}=(1.00g/mL\times 3.70mL)=3.70g$

To calculate the percentage yield of water, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of water = 3.70 g

Theoretical yield of water = 4.27 g

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{3.70g}{4.27g}\times 100\\\\\% \text{yield of water}=86.6\%$

Hence, the percentage yield of water is 86.6 %