How much bromobenzene (C6H5Br) can we make from 6.58 g of benzene (C6H6) and 3.81 g of bromine according to the equation below?
You didn't give an equation.
To determine how much bromobenzene can be produced from the given amounts of benzene and bromine, we first need to balance the chemical equation.
The balanced chemical equation for the reaction between benzene and bromine to form bromobenzene is:
C6H6 + Br2 -> C6H5Br + HBr
According to the equation, 1 mole of benzene reacts with 1 mole of bromine to produce 1 mole of bromobenzene.
To find the number of moles of benzene and bromine given in the question, we can use the formula:
moles = mass / molar mass
The molar masses of benzene (C6H6) and bromine (Br2) can be found using the periodic table:
C6H6: (6 * 12.011 g/mol) + (6 * 1.008 g/mol) = 78.114 g/mol
Br2: 2 * 79.904 g/mol = 159.808 g/mol
Let's calculate the number of moles of benzene and bromine:
moles of benzene = 6.58 g / 78.114 g/mol
moles of bromine = 3.81 g / 159.808 g/mol
Now, we can determine the limiting reagent, which is the reactant that will be completely consumed and determine the maximum amount of product that can be formed.
To determine the limiting reagent, we compare the moles of benzene and bromine. Whichever is present in a lesser amount will be the limiting reagent.
Assuming the moles of benzene is "x" and the moles of bromine is "y", we have:
x = moles of benzene
y = moles of bromine
If x < y, then benzene is the limiting reagent.
If x > y, then bromine is the limiting reagent.
If x = y, then both reactants are in stoichiometric amounts, and there is no limiting reagent.
Now, let's calculate the moles of benzene and bromine:
moles of benzene = 6.58 g / 78.114 g/mol = 0.0842 mol
moles of bromine = 3.81 g / 159.808 g/mol = 0.0238 mol
Since x > y, bromine is the limiting reagent.
To find the moles of bromobenzene produced, we can use the stoichiometry of the balanced equation. From the balanced equation, we can see that 1 mole of benzene reacts with 1 mole of bromine to produce 1 mole of bromobenzene.
Therefore, the moles of bromobenzene formed will be equal to the moles of bromine consumed. In this case, it is 0.0238 mol.
Finally, to determine the mass of bromobenzene produced, we can use the formula:
mass = moles * molar mass
The molar mass of bromobenzene (C6H5Br) can be calculated as follows:
C6H5Br: (6 * 12.011 g/mol) + (5 * 1.008 g/mol) + 79.904 g/mol = 157.016 g/mol
mass of bromobenzene = 0.0238 mol * 157.016 g/mol = 3.73 g
Therefore, 3.73 grams of bromobenzene can be produced from 6.58 grams of benzene and 3.81 grams of bromine based on the balanced equation and stoichiometry.