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.
<u>Answer:</u> The percent yield of the reaction is 90.71 %.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For bromine gas:</u>
Given mass of bromine gas = 16.1 g
Molar mass of bromine gas = 159.8 g/mol
Putting values in equation 1, we get:
- <u>For chlorine gas:</u>
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 %.