Nitrobenzene, C6H5NO2, is prepared from benzene, C6H6, and nitric acid, HNO3. The other product of the reaction is water. When 22.4 g of benzene is reacted with an excess of nitric acid, 31.6 g of nitrobenzene is produced. What is the percent yield for this reaction?

1. Write and balanced the equation.

2. Convert 22.4 g benzene to mols. mols = grams/molar mass = ?
3. Using the coefficients in the balanced equation, convert mols benzene to mols nitrobenzene. This is the theoretical (TY) for the reaction.
4. The actual yield (AY) is 31.6
5. %yield = (AY/TY)*100 = ?

answer

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Ah, chemistry! The ultimate mix-n-match game. Let's break down the question and calculate that percent yield, shall we?

Given:
Mass of benzene = 22.4 g
Mass of nitrobenzene produced = 31.6 g

First, we need to determine the theoretical yield, which is the maximum amount of nitrobenzene that can be produced based on the given reactants.

The molecular weight of benzene (C6H6) is 78.11 g/mol.
The molecular weight of nitrobenzene (C6H5NO2) is 123.11 g/mol.

To figure out how many moles of benzene reacted, we divide the mass by the molecular weight:
moles of benzene = 22.4 g / 78.11 g/mol

Since benzene and nitrobenzene have a 1:1 ratio in the balanced equation, this means moles of nitrobenzene produced = moles of benzene reacted.

So, the theoretical yield of nitrobenzene would be:
moles of nitrobenzene = 22.4 g / 78.11 g/mol

Now we can calculate the mass of the theoretical yield:
mass of theoretical yield = moles of nitrobenzene * molecular weight of nitrobenzene

Finally, we can calculate the percent yield:
percent yield = (actual yield / theoretical yield) * 100

Hope that wasn't too dry for you! Let's crunch those numbers and find out the percent yield!

To calculate the percent yield for this reaction, we need to compare the actual yield (the amount of nitrobenzene obtained in the reaction) with the theoretical yield (the maximum amount of nitrobenzene that could be obtained based on stoichiometry).

First, let's determine the molecular weight of benzene (C6H6), nitrobenzene (C6H5NO2), and nitric acid (HNO3):

- Molecular weight of benzene (C6H6): 6(12.01 g/mol) + 6(1.01 g/mol) = 78.11 g/mol
- Molecular weight of nitrobenzene (C6H5NO2): 6(12.01 g/mol) + 5(1.01 g/mol) + 1(14.01 g/mol) + 2(16.00 g/mol) = 123.11 g/mol
- Molecular weight of nitric acid (HNO3): 1(1.01 g/mol) + 1(14.01 g/mol) + 3(16.00 g/mol) = 63.01 g/mol

Next, we calculate the theoretical yield of nitrobenzene using stoichiometry. The balanced chemical equation for the reaction is:

C6H6 + HNO3 -> C6H5NO2 + H2O

From the balanced equation, we see that the ratio of benzene to nitrobenzene is 1:1. Therefore, the amount (in moles) of nitrobenzene produced is equal to the amount of benzene used.

Converting the mass of benzene to moles:
moles of benzene = 22.4 g benzene / 78.11 g/mol = 0.2862 mol benzene

Since the ratio of benzene to nitrobenzene is 1:1, the moles of nitrobenzene produced will also be 0.2862 mol.

Now, we can calculate the theoretical yield of nitrobenzene in grams:
theoretical yield = moles of nitrobenzene × molecular weight of nitrobenzene
theoretical yield = 0.2862 mol × 123.11 g/mol = 35.34 g

Finally, we can calculate the percent yield using the following formula:
percent yield = (actual yield / theoretical yield) × 100

Given that the actual yield is 31.6 g of nitrobenzene, we can substitute the values into the formula:
percent yield = (31.6 g / 35.34 g) × 100 = 89.5%

Therefore, the percent yield for this reaction is 89.5%.

Consider the preparation of nitrobenzene, C6H5NO2, by the reaction of a limited amount of benzene,

C6H6, with excess nitric acid, HNO3. The balanced equation for the reaction may be written as
C6H6 + HNO3 → C6H5NO2 + H2O
A 15.6-g sample of C6H6 is mixed with excess HNO3. We isolate 18.0 g of C6H5NO2. What is the
percent yield of C6H5NO2 in this reaction?