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.
Answer: 72.2 g
