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 .

The percent yield of ethyl chloride is 83.03 %

Explanation:

To calculate the number of moles, we use the equation

  ....(1)

Given mass of ethane gas = 300 g

Molar mass of ethane gas = 30 g/mol

Putting values in equation 1, we get:

Given mass of chlorine gas = 650 g

Molar mass of chlorine gas = 71 g/mol

Putting values in equation 1, we get:

For the given chemical equation:

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 = 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 = 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:

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

Theoretical yield of ethyl chloride = 590.175 g

Experimental yield of ethyl chloride = 490 g

Putting values in above equation, we get:

Hence, the percent yield of ethyl chloride is 83.03 %