Problem: 40.0 g of C6H6 are allowed to react with 75.0 g of Br2 according to the following balanced equation: C6H6 + Br2 → C6H5Br + HBra. How many moles of C6H5Br can be produced from 40.0 g C6H6?b. How many moles of C6H5Br can be produced from 75.0 g Br2?c. What is the limiting reactant? (WHich one makes the least moles of C6H5Br)d. What is the theoretical yield of C6H5Br in grams?e. If the actual yield C6H5Br was 59.7 g, what is the percent yield?

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How many moles of C6H5Br can be produced from 40.0 g C6H6?


C6H6 + Br2 → C6H5Br + HBr

The flow for this problem will be like this:


Mass C6H6 (molar mass C6H6 )→ Moles of C6H6 (mole-to-mole comparison)  Moles of C6H5Br 


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Problem Details

40.0 g of C6H6 are allowed to react with 75.0 g of Br2 according to the following balanced equation: 


C6H6 + Br2 → C6H5Br + HBr

a. How many moles of C6H5Br can be produced from 40.0 g C6H6?

b. How many moles of C6H5Br can be produced from 75.0 g Br2?

c. What is the limiting reactant? (WHich one makes the least moles of C6H5Br)

d. What is the theoretical yield of C6H5Br in grams?

e. If the actual yield C6H5Br was 59.7 g, what is the percent yield?

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