16.1g of bromine are mixed with 8.42g of chlorite to give an actual yield of 21.1g of bromine monochloride. Solve using factor label.

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16.1 g of bromine are mixed with 8.42g of chlorine to give an actual yield of 21.1 g of bromine monochloride. Determine the percent yield of the reaction.

The percent yield of the reaction is 90.71 %.

Explanation:

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

    .....(1)

Given mass of bromine gas = 16.1 g

Molar mass of bromine gas = 159.8 g/mol

Putting values in equation 1, we get:

Given mass of chlorine gas = 8.42 g

Molar mass of chlorine gas = 71 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction of bromine and chlorine gas follows:

By Stoichiometry of the reaction:

1 mole of bromine gas reacts with 1 mole of chlorine gas

So, 0.1008 moles of bromine gas will react with = of chlorine gas

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

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

By Stoichiometry of the reaction:

1 mole of bromine gas produces 2 mole of bromine monochloride

So, 0.1008 moles of bromine gas will produce = of bromine monochloride

Now, calculating the mass of bromine monochloride from equation 1, we get:

Molar mass of bromine monochloride = 115.36 g/mol

Moles of bromine monochloride = 0.2016 moles

Putting values in equation 1, we get:

To calculate the percentage yield of bromine monochloride, we use the equation:

Actual yield of bromine monochloride = 21.1 g

Theoretical yield of bromine monochloride = 23.26 g

Putting values in above equation, we get:

Hence, the percent yield of the reaction is 90.71 %.