Question

The reaction of ethane gas (C2H6) with chlorine gas (Cl2) produces C2H5Cl as its main product. Calculate the percent yield of C2H5Cl if the reaction of 300 g of ethane with 650 g of chlorine produced 490 g of C2H5Cl .

Answer 1

Answer: The percent yield of ethyl chloride is 83.03 %

Explanation:

To calculate the number of moles, we use the equation

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$   ....(1)

• For ethane gas:

Given mass of ethane gas = 300 g

Molar mass of ethane gas = 30 g/mol

Putting values in equation 1, we get:

$\text{Moles of ethane gas}=\frac{300g}{30g/mol}=10mol$

• For chlorine gas:

Given mass of chlorine gas = 650 g

Molar mass of chlorine gas = 71 g/mol

Putting values in equation 1, we get:

$\text{Moles of chlorine gas}=\frac{650g}{71g/mol}=9.15mol$

For the given chemical equation:

$C_2H_6(g)+Cl_2(g)\rightarrow C_2H_5Cl(s)+HCl(g)$

By Stoichiometry of the reaction:

1 mole of chlorine gas reacts with 1 moles of ethane gas.

So, 9.15 moles of chlorine gas will react with = $\frac{1}{1}\times 9.15=9.15moles$ of ethane gas.

As, given amount of ethane gas is more than the required amount. So, it is considered as an excess reagent.

Thus, chlorine gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of chlorine gas produces 1 mole of ethyl chloride.

So, 9.15 moles of chlorine gas will produce = $\frac{1}{1}\times 9.15=9.15moles$ of ethyl chloride

Now, calculating the mass of ethyl chloride from equation 1, we get:

Molar mass of ethyl chloride = 64.5 g/mol

Moles of ethyl chloride = 9.15 moles

Putting values in equation 1, we get:

$9.15mol=\frac{\text{Mass of ethyl chloride}}{64.5g/mol}\\\\\text{Mass of ethyl chloride}=590.175g$

To calculate the percent yield of ethyl chloride, we use the equation:

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

Theoretical yield of ethyl chloride = 590.175 g

Experimental yield of ethyl chloride = 490 g

Putting values in above equation, we get:

$\%\text{ yield of ethyl chloride}=\frac{490g}{590.175g}\times 100\\\\\%\text{ yield of ethyl chloride}=83.03\%$

Hence, the percent yield of ethyl chloride is 83.03 %