we're given the equation:
C6H6(l)+Br2(l) --> C6H5Br(l)+HBr(g)
you need to prepare 50.0g pf bromobenzene and you expect no more than a 75% yield. How much benzene should you begin with if the yield is 75%.
i know that % yield = (actual/theoretical) * 100 %
but i don't know how to get the actual/theoretical values. would 50.0 g be put as an actual or theoretical value? The way i understand it, is that it should be used as the actual value because that is the amount you want to prepare, and if its put as a theoretical value the actual amount would always be less than theoretical. Am i thinking right?
This question confuses me, maybe it confuses me because they say 75% twice and it throws me off lol. please explain steps as well, thank you in advance!
To calculate the amount of benzene you should begin with if the yield is 75%, you need to find the theoretical yield first. The theoretical yield is the maximum amount of product that can be obtained based on the stoichiometry of the balanced equation.
Let's start by calculating the molar mass of bromobenzene (C6H5Br):
- The molar mass of carbon (C) is 12.01 g/mol.
- The molar mass of hydrogen (H) is 1.01 g/mol.
- The molar mass of bromine (Br) is 79.90 g/mol.
So, the molar mass of bromobenzene (C6H5Br) is calculated as follows:
6(12.01 g/mol) + 6(1.01 g/mol) + 79.90 g/mol = 157.13 g/mol.
Now, we can calculate the theoretical yield using the given information:
- The desired amount of bromobenzene is 50.0 g.
- The % yield is 75%.
The formula for calculating the theoretical yield is:
Theoretical Yield = (Actual Yield / % Yield) * 100
Let's substitute the values into the formula:
Theoretical Yield = (50.0 g / 75%) * 100 = (50.0 g / 0.75) * 100 = 66.67 g (approx.)
Therefore, the theoretical yield of bromobenzene is approximately 66.67 g.
Now, we can find the amount of benzene needed to achieve this theoretical yield. The stoichiometry of the balanced equation tells us that the reaction is 1:1 between benzene and bromobenzene. This means that for every 1 mole of bromobenzene produced, 1 mole of benzene reacts. Therefore, the molar mass of benzene is the same as the molar mass of bromobenzene.
Using the molar mass of benzene (157.13 g/mol), we can calculate the amount of benzene needed:
Amount of benzene = Theoretical Yield of bromobenzene = 66.67 g.
Therefore, you should begin with approximately 66.67 g of benzene to obtain a 75% yield of bromobenzene.
To solve this problem, we need to determine the amount of benzene we should start with in order to produce 50.0g of bromobenzene, assuming a maximum yield of 75%.
In a chemical reaction, the theoretical yield is the maximum amount of product that can be obtained based on the stoichiometry of the balanced equation. The actual yield, on the other hand, is the amount of product obtained in a real experiment.
In this case, we want to prepare 50.0g of bromobenzene, which is the desired actual yield.
To calculate the theoretical yield, we need to use the balanced equation:
C6H6(l) + Br2(l) --> C6H5Br(l) + HBr(g)
From the equation, we can see that the molar ratio between benzene (C6H6) and bromobenzene (C6H5Br) is 1:1. This means that for every 1 mole of benzene, we can obtain 1 mole of bromobenzene.
To calculate the theoretical amount of benzene needed, we can convert the 50.0g of bromobenzene to moles, and then use the molar ratio to find the moles of benzene required. Finally, we can convert the moles of benzene to grams.
Here are the steps:
Step 1: Convert the desired actual yield of bromobenzene to moles.
50.0g bromobenzene x (1 mole/ molar mass of bromobenzene) = moles of bromobenzene
Step 2: Use the molar ratio to find the moles of benzene required.
moles of bromobenzene x (1 mole of benzene/ 1 mole of bromobenzene) = moles of benzene
Step 3: Convert moles of benzene to grams.
moles of benzene x (molar mass of benzene/ 1 mole of benzene) = grams of benzene
Note: The molar mass of benzene (C6H6) is 78.11 g/mol, and the molar mass of bromobenzene (C6H5Br) is 157.01 g/mol.
So by following these steps, you can calculate the amount of benzene needed to prepare 50.0g of bromobenzene with an expected yield of 75%.
This is a stoichiometry problem with an extra step of % thrown into the mix.
C6H6 + Br2 ==> C6H5Br + HBr.
mols bromobenzene in 50 g is 50/molar mass bromobenzene.
Now you wan to convert mols bromobenzene to mols benzene. Since the equation shows you it is 1 mol benzene to 1 mol bromobenzene, then mols benzene = mols bromobenzene.
Now convert mols benzene to grams. That's g = mols x molar mass.
That gives you grams benzene you would need to start with to prepare 50.0g bromobenzene (that's what the problem wanted) BUT that is for the theoretical yield which is 100%. Let's call grams benzene you've calculated something like n.
I find it easy to say, "What number grams benzene x 0.75 = n grams benzene?" In equation form that is
? g benzene x 0.75 = n g. Solve for ?g benzene.