In an experiment a student allowed benzene C6H6+ TO REACT WITH EXCESS BROMINE BR2+ In an attempt to prepare bromobenzene C6H5Br. This reaction also produced as a by-product, dibromobenzene, C6H4Br2+ On the basis of the equation.

What is the maximum amount of C6H5Br that the student could have hoped to obtain from 15.0 g of benzene? (This is the theoretical yield.)

I don't get the C6H6+, Br2+ and C6H4B42+. Why the + signs?

You don't have them in the reaction.
Convert 15.0 g benzene to mols. mols = grams/molar mass.
Use the coefficients in the balanced equation to convert mols benzene to mols of the bromobenzene.
Now convert these mols to grams. g = mols bromobenzene x molar mass benzene.

To determine the maximum amount of C6H5Br that the student could have obtained, we need to calculate the theoretical yield using the balanced equation:

C6H6 + Br2 -> C6H5Br + HBr

The molar mass of benzene (C6H6) is 78 g/mol, and the molar mass of bromobenzene (C6H5Br) is 157 g/mol.

Step 1: Calculate the number of moles of benzene (C6H6) in 15.0 g.
Number of moles = mass / molar mass
Number of moles of benzene = 15.0 g / 78 g/mol = 0.192 mol

Step 2: Based on the balanced equation, the stoichiometric ratio between benzene and bromobenzene is 1:1. Therefore, the maximum amount of C6H5Br that can be obtained is also 0.192 mol.

Step 3: Calculate the maximum amount of C6H5Br in grams.
Mass of C6H5Br = number of moles × molar mass
Mass of C6H5Br = 0.192 mol × 157 g/mol = 30.144 g

The maximum amount of C6H5Br that the student could have hoped to obtain from 15.0 g of benzene is 30.144 g.

To determine the maximum amount of C6H5Br (bromobenzene) that the student could have obtained, we need to calculate the theoretical yield. This is done by stoichiometry, using the balanced chemical equation provided.

The balanced equation for the reaction is:
C6H6 + 3Br2 → C6H5Br + BrC6H4Br

From the equation, we can see that 1 mole of benzene (C6H6) reacts with 3 moles of bromine (Br2) to produce 1 mole of bromobenzene (C6H5Br), as well as 1 mole of dibromobenzene (BrC6H4Br).

To find the theoretical yield of C6H5Br, we'll follow these steps:

Step 1: Calculate the molar mass of benzene (C6H6) and bromobenzene (C6H5Br).
- Molar mass of C6H6 = (6 * atomic mass of C) + (6 * atomic mass of H)
= (6 * 12.01 g/mol) + (6 * 1.01 g/mol)
= 78.11 g/mol
- Molar mass of C6H5Br = (6 * atomic mass of C) + (5 * atomic mass of H) + atomic mass of Br
= (6 * 12.01 g/mol) + (5 * 1.01 g/mol) + 79.90 g/mol
= 157.02 g/mol

Step 2: Calculate the number of moles of benzene (C6H6) in 15.0 g.
- Moles of C6H6 = Mass of C6H6 / Molar mass of C6H6
= 15.0 g / 78.11 g/mol
≈ 0.192 mol

Step 3: Based on the stoichiometric ratio, determine the number of moles of C6H5Br produced.
- Since the reaction ratio of C6H6 to C6H5Br is 1:1, the number of moles of C6H5Br is also 0.192 mol.

Step 4: Calculate the mass of C6H5Br produced.
- Mass of C6H5Br = Moles of C6H5Br * Molar mass of C6H5Br
= 0.192 mol * 157.02 g/mol
≈ 30.15 g

Therefore, the maximum amount of C6H5Br that the student could have hoped to obtain from 15.0 g of benzene (C6H6) is approximately 30.15 g (theoretical yield).