What happens to the particles in a substance as it freezes?

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 l

krok68 [10]

The question is incomplete, here is the complete question:

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.

Answer: The percent yield of the reaction is 90.71 %.

Explanation:

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

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

For bromine gas:

Given mass of bromine gas = 16.1 g

Molar mass of bromine gas = 159.8 g/mol

Putting values in equation 1, we get:

$\text{Moles of bromine gas}=\frac{16.1g}{159.8g/mol}=0.1008mol$

For chlorine gas:

Given mass of chlorine gas = 8.42 g

Molar mass of chlorine gas = 71 g/mol

Putting values in equation 1, we get:

$\text{Moles of chlorine gas}=\frac{8.42g}{71g/mol}=0.118mol$

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

$Br_2+Cl_2\rightarrow 2BrCl$

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 = $\frac{1}{1}\times 0.1008=0.1008mol$ 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 = $\frac{2}{1}\times 0.1008=0.2016moles$ 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:

$0.2016mol=\frac{\text{Mass of bromine monochloride}}{115.36g/mol}\\\\\text{Mass of bromine monochloride}=(0.2016mol\times 115.36g/mol)=23.26g$

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

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

Actual yield of bromine monochloride = 21.1 g

Theoretical yield of bromine monochloride = 23.26 g

Putting values in above equation, we get:

$\%\text{ yield of bromine monochloride}=\frac{21.1g}{23.26g}\times 100\\\\\% \text{yield of bromine monochloride}=90.71\%$

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

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