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

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.

To determine the limiting reagent, we need to compare the amount of benzene and bromine used in the reaction.

The molar mass of benzene (C6H6) is:
C = 12.01 g/mol
H = 1.01 g/mol
6C + 6H = 6(12.01) + 6(1.01) = 78.11 g/mol

The molar mass of bromine (Br2) is:
2Br = 2(79.9) = 159.8 g/mol

Using the molar masses, we can calculate the number of moles of each substance:

Number of moles of benzene:
60.0 g / 78.11 g/mol = 0.767 mol

Number of moles of bromine:
135 g / 159.8 g/mol = 0.845 mol

To determine the limiting reagent, we compare the mole ratios of benzene and bromine based on the balanced chemical equation:

C6H6 + Br2 -> C6H5Br + HBr

From the equation, we can see that 1 mole of benzene reacts with 1 mole of bromine. Comparing the moles of benzene (0.767 mol) and bromine (0.845 mol), we can see that benzene is the limiting reagent because it is present in the smaller amount.

To calculate the theoretical yield, we need to convert the moles of benzene to moles of bromobenzene (C6H5Br) using the stoichiometric ratio:

From the equation: 1 mole of benzene reacts to give 1 mole of bromobenzene.

The moles of bromobenzene can be calculated as:
0.767 mol of benzene x (1 mol of bromobenzene / 1 mol of benzene) = 0.767 mol of bromobenzene

Finally, we can calculate the theoretical yield of bromobenzene in grams by multiplying the moles of bromobenzene by its molar mass:

Molar mass of bromobenzene (C6H5Br):
C = 12.01 g/mol
H = 1.01 g/mol
Br = 79.9 g/mol

Molar mass = 12.01 + 5(1.01) + 79.9 = 157.07 g/mol

Theoretical yield = 0.767 mol x 157.07 g/mol = 119.38 g

Therefore, the theoretical yield of bromobenzene is 119.38 grams.

To determine the limiting reagent and theoretical yield in a chemical reaction, we need to compare the amounts of each reactant to the stoichiometry of the balanced equation.

First, let's calculate the number of moles for benzene (C6H6) and bromine (Br2) using their molar masses. The molar mass of benzene is 78.11 g/mol, and the molar mass of bromine is 159.81 g/mol.

The number of moles of benzene (C6H6) can be calculated using the formula:

moles = mass / molar mass
moles of C6H6 = 60.0 g / 78.11 g/mol

Calculating this, we find moles of C6H6 ≈ 0.767 mol.

Similarly, the number of moles of bromine (Br2) can be calculated as:

moles of Br2 = 135 g / 159.81 g/mol

Calculating this, we find moles of Br2 ≈ 0.845 mol.

Now, let's determine the limiting reagent. The stoichiometric ratio of benzene to bromine in the balanced equation is 1:1. This means one mole of benzene reacts with one mole of bromine.

From the mole calculations above, we have 0.767 moles of C6H6 and 0.845 moles of Br2.

Since the ratio of C6H6 to Br2 is 1:1, we can see that there is an excess of bromine. Therefore, the limiting reagent is benzene (C6H6).

To find the theoretical yield, we need to consider the balanced equation from the reaction:

C6H6 + Br2 → C6H5Br + HBr

The coefficient of bromobenzene (C6H5Br) in the balanced equation is 1, meaning that 1 mole of benzene (C6H6) will produce 1 mole of bromobenzene (C6H5Br).

Since benzene is the limiting reagent, the moles of bromobenzene (C6H5Br) produced will be equal to the moles of benzene (C6H6). Therefore, the theoretical yield of bromobenzene can be calculated as follows:

theoretical yield = moles of C6H5Br = moles of C6H6

Substituting the value of moles of C6H6 we calculated earlier:

theoretical yield ≈ 0.767 mol

Please note that the theoretical yield represents the maximum amount of product that can be obtained under ideal conditions and assumes the reaction goes to completion without any side reactions or losses.