When benzene (C6H6) reacts with bromine (Br2), bromobenzene (C6H5Br) is obtained:
C6H6 + Br2 → C6H5Br + HBr
A.) What is the theoretical yield of bromobenzene in this reaction when 34.0 g of benzene reacts with 73.7g of bromine?
B.) If the actual yield of bromobenzene was 65.6 g, what was the percentage yield?
C6H6 + Br2 → C6H5Br + HBr
A.) What is the theoretical yield of bromobenzene in this reaction when 34.0 g of benzene reacts with 73.7g of bromine?
B.) If the actual yield of bromobenzene was 65.6 g, what was the percentage yield?
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These problems are not that hard if you learn the 4 steps involved (to go from a mass of one thing to mass of another).
The four steps are:
Mass reactant(s) [usually given to you in the problem] --> moles of reactant(s) --> moles of product [the one you are asked to find in the problem] --> mass of product
You are told you have 34.0g of benzene and 73.7g of bromine. These are your two reactants. You must determine which of these is the limiting reactant. I like to think of the limiting reactant like a recipe when you are cooking. If your recipe calls for 2 eggs and 1 cup of milk for a cake... you can continue to make cake as long as you have enough of both. However, if one ingredient runs out before the other, you can no longer make cake.
Think of the balanced chemical equation as a recipe. You need 1 mole of benzene, and 1 mole of bromine to form 1 mole of bromobenzene and hydrogen bromide.
So the best thing to do is determine how many moles of product would be form if each of the reactants were completely consumed. Whichever reactant forms the least amount of product, is your limiting reagent.
34.0g of benzene has a molar mass of 78.108g.
34.0g benzene x (1 mole benzene/78.108g benzene) x (1 mole C6H5Br/1 mole benzene) = 0.435247201 mol C6H5Br
73.7g of bromine (Br2) has a molar mass of 159.808g.
73.7g bromine x (1 mole bromine/159.808g) x (1 mole C6H5Br/1 mole bromine) = 0.4611784141 mol Br2
Since we have less product formed when all of the benzene reacts, benzene is our limiting reactant. It determine how much of the products that we can form. We have more bromine than we need.
.4352947201 mol bromine x (157.004g C6H5Br/1 mol C6H5Br) =68.34301229g C6H5Br.
'Theoretically' you can form a maximum of 68.34301229g of C6H5Br. This is the answer to part A.
Now for part B, you use the theoretical yield (the yield you should get based on mathematical data), and figure out what percentage of the yield actually received.
The yield was 65.6g. The percentage is 65.6/68.34 = .959906 x 100% = 96.0%.
Think of percent yield as a test. On an exam, theoretically, you should get a 100%. But this never happens, so you receive your grade based on percentage you got (actual/theoretical).
Sorry for the long winded answer, I believe in teaching more so than giving answers.
The four steps are:
Mass reactant(s) [usually given to you in the problem] --> moles of reactant(s) --> moles of product [the one you are asked to find in the problem] --> mass of product
You are told you have 34.0g of benzene and 73.7g of bromine. These are your two reactants. You must determine which of these is the limiting reactant. I like to think of the limiting reactant like a recipe when you are cooking. If your recipe calls for 2 eggs and 1 cup of milk for a cake... you can continue to make cake as long as you have enough of both. However, if one ingredient runs out before the other, you can no longer make cake.
Think of the balanced chemical equation as a recipe. You need 1 mole of benzene, and 1 mole of bromine to form 1 mole of bromobenzene and hydrogen bromide.
So the best thing to do is determine how many moles of product would be form if each of the reactants were completely consumed. Whichever reactant forms the least amount of product, is your limiting reagent.
34.0g of benzene has a molar mass of 78.108g.
34.0g benzene x (1 mole benzene/78.108g benzene) x (1 mole C6H5Br/1 mole benzene) = 0.435247201 mol C6H5Br
73.7g of bromine (Br2) has a molar mass of 159.808g.
73.7g bromine x (1 mole bromine/159.808g) x (1 mole C6H5Br/1 mole bromine) = 0.4611784141 mol Br2
Since we have less product formed when all of the benzene reacts, benzene is our limiting reactant. It determine how much of the products that we can form. We have more bromine than we need.
.4352947201 mol bromine x (157.004g C6H5Br/1 mol C6H5Br) =68.34301229g C6H5Br.
'Theoretically' you can form a maximum of 68.34301229g of C6H5Br. This is the answer to part A.
Now for part B, you use the theoretical yield (the yield you should get based on mathematical data), and figure out what percentage of the yield actually received.
The yield was 65.6g. The percentage is 65.6/68.34 = .959906 x 100% = 96.0%.
Think of percent yield as a test. On an exam, theoretically, you should get a 100%. But this never happens, so you receive your grade based on percentage you got (actual/theoretical).
Sorry for the long winded answer, I believe in teaching more so than giving answers.