What is the theoretical yield of c6h5br if 42.1 g of c6h6 react with 73.0 g of br2?

1) Chemical reaction

C6H6 + Br2 ---> C6H5Br + HBr

2) Molar raios

1 mol C6H6 : 1 mol Br2 : 1 mol C6H5Br

3) Convert the data to moles

42.1 g of C6H6

molar mass of C6H6 = 6*12g/mol + 6*1g/mol = 78 g/mol

42.1 g / 78 g/mol = 0.54 mol

73.0 g of Br2

molar mass of Br2 = 2 * 79.9 g/mol = 159.8 g/mol

73.0 g / 159.8 g/mol = 0.46 mol of Br2

=> limiting reagent is the Br2.

4) Product

1 mol of Br2 (limiting reagent) yields 1 mole of C6H5Br, then you will obtain 0.46 mol of C6H5Br

5) Convert the product to grams

molar mass of C6H5Br = 6*12g/mol + 5*1g/mol + 79.9 g/mol = 156.9 g/mol

0.46mol*156.9g/mol = 72.2 g.

72.2 g