60.0 g of benzene is mixed with 135g of bromine to form bromobenzene according to the following reactionC6H6+Br2-C6H5Br+HBr which is the limiting reagent and what is the theoretical yield

1 answer

Convert 60.0 g benzene to moles. moles = grams/molar mass.
Convert 136 g Br2 t moles the same way.

Using the coefficients in the balanced equation (it's 1:1), convert moles benzene to moles of the product.
Same procedure, convert moles Br2 to moles of the product.
These numbers probably will be different which means that one of them is wrong. The correct answer, in limiting reagent problems, is ALWAYS the smaller one and the reagent producing that value is the limiting reagent.

Finally, using the smaller value from above, convert moles of the product to grams. g = moles x molar mass.