Consider the following reaction:

C6H5CL+C2HOCL3=>C14H9CL5+H20

If 1142 g of C6H5CL is reacted with 485 g of C2HOCL3, what mass of C14H9CL5 would be formed?

I believe I told you how to do this problem yesterday. If you have any questions about the procedure, show your work and we can work it out together.

To determine the mass of C14H9CL5 formed in the reaction, we first need to find out which reactant is limiting.

To do this, we calculate the number of moles of each reactant using their molar masses. The molar masses are as follows:

Molar mass of C6H5CL = 112.56 g/mol
Molar mass of C2HOCL3 = 163.82 g/mol

Now we can calculate the moles of each reactant:

Moles of C6H5CL = mass / molar mass = 1142 g / 112.56 g/mol
Moles of C2HOCL3 = mass / molar mass = 485 g / 163.82 g/mol

Next, we need to determine the stoichiometry of the reaction. According to the balanced equation, 1 mole of C6H5CL reacts with 1 mole of C2HOCL3 to produce 1 mole of C14H9CL5.

Now let's compare the moles of reactants to find the limiting reactant.

Moles of C6H5CL = 1142 g / 112.56 g/mol = 10.14 mol
Moles of C2HOCL3 = 485 g / 163.82 g/mol = 2.96 mol

We can see that there are fewer moles of C2HOCL3 compared to C6H5CL. Therefore, C2HOCL3 is the limiting reactant.

To find the mass of C14H9CL5 formed, we need to use the mole ratio between C2HOCL3 and C14H9CL5 from the balanced equation, which is 1:1. So, the moles of C14H9CL5 formed will be equal to the moles of C2HOCL3.

Moles of C14H9CL5 = 2.96 mol

Finally, we can calculate the mass of C14H9CL5 formed using its molar mass:

Mass of C14H9CL5 = Moles of C14H9CL5 x Molar mass of C14H9CL5
= 2.96 mol x (596.30 g/mol)
= 1764.78 g

Therefore, the mass of C14H9CL5 formed in the reaction would be approximately 1764.78 grams.

To find the mass of C14H9CL5 formed, we need to use the given information about the reaction and the reactants' molar masses.

Step 1: Calculate the number of moles of each substance.

The molar mass of C6H5CL (chlorobenzene) is approximately 112.57 g/mol.
Number of moles of C6H5CL = Mass of C6H5CL / Molar mass of C6H5CL

Number of moles of C6H5CL = 1142 g / 112.57 g/mol

Now, let's calculate the number of moles of C2HOCL3 (chloral) using its molar mass, which is approximately 165.36 g/mol.

Number of moles of C2HOCL3 = Mass of C2HOCL3 / Molar mass of C2HOCL3

Number of moles of C2HOCL3 = 485 g / 165.36 g/mol

Step 2: Determine the limiting reactant.

To determine the limiting reactant, we compare the mole ratios of the reactants based on the balanced equation coefficients. The reactant that produces the least amount of product is the limiting reactant.

From the balanced equation:

1 mol C6H5CL + 1 mol C2HOCL3 → 1 mol C14H9CL5

The mole ratio of C6H5CL to C2HOCL3 is 1:1.

The moles of C6H5CL and C2HOCL3 calculated in Step 1 represent the actual moles of the respective reactants used in the reaction.

Since the mole ratio is 1:1, the limiting reactant is the reactant that has fewer moles.

Compare the moles of C6H5CL and C2HOCL3. The reactant with fewer moles is the limiting reactant.

Step 3: Calculate the moles of C14H9CL5 formed.

Since the limiting reactant is C6H5CL, the moles of C14H9CL5 formed will be equal to the moles of C6H5CL.

Step 4: Convert moles of C14H9CL5 to grams.

To find the mass of C14H9CL5 formed, we use the molar mass of C14H9CL5, which is approximately 364.77 g/mol.

Mass of C14H9CL5 = Moles of C14H9CL5 × Molar mass of C14H9CL5

Now let's calculate:

Mass of C14H9CL5 = Moles of C6H5CL × Molar mass of C14H9CL5

Finally, plug in the value of moles of C6H5CL (calculated in Step 3):

Mass of C14H9CL5 = Moles of C6H5CL × Molar mass of C14H9CL5

Note: The molar masses and balanced equation coefficients used in this calculation are approximate and rounded values.