DDT, an insecticide harmful to fish, birds, and humans, is produced by the following reaction:

2 C 6 H 5 Cl (Chlorobenzene) + C 2 H O Cl 3(chloral) ==> C 14 H 9 Cl 5 (DDT) + H 2 O

In a government lab, 1142 g of chlorobenzene is reacted with 485g of Chloral.

A)What mass of DDT is formed?
B)Which reactant is limitin? Which is in excess?
C)What mass of the excess reactant is left over?
D)If the actual yield of DDT is 200.0 g, what is the percent yield?

Please be specific (in showing work), and show significant figures! Thank you!

You have the equation. Make sure it is balanced.

Convert g chlorobenzene to moles. moles = grams/molar mass.
Convert g chloral to moles.

Using the coefficients in the balanced equation, convert moles chlorobenzene to moles DDT.
Same procedure, convert moles chloral to moles DDT.
More than likely, the numbers for moles DDT will not be the same which means one of them is wrong. In limiting reagent problems, the correct value is ALWAYS the smaller one and the reactant providing that number is the limiting reagent.

Now convert the smaller value to grams. g = mols x molar mass.

A) To find the mass of DDT formed, we need to determine the limiting reactant and calculate the mass of DDT produced using stoichiometry.

First, let's find the number of moles of each reactant:

The molar mass of chlorobenzene (C6H5Cl) = (12.01 g/mol x 6) + (1.01 g/mol x 5) + 35.45 g/mol = 112.56 g/mol
Number of moles of chlorobenzene = mass / molar mass = 1142 g / 112.56 g/mol = 10.14 mol

The molar mass of chloral (C2HCl3O) = 119.38 g/mol
Number of moles of chloral = mass / molar mass = 485 g / 119.38 g/mol = 4.06 mol

According to the balanced equation, the stoichiometric ratio between chlorobenzene and DDT is 2:1. Therefore, the molar ratio between chlorobenzene and DDT is 2:1.

Since the molar ratio of chlorobenzene to DDT is 2:1, the number of moles of DDT formed will be half of the moles of chlorobenzene used.

Number of moles of DDT formed = 10.14 mol / 2 = 5.07 mol

Now, let's calculate the mass of DDT formed:

The molar mass of DDT (C14H9Cl5) = (12.01 g/mol x 14) + (1.01 g/mol x 9) + (35.45 g/mol x 5) = 354.49 g/mol
Mass of DDT formed = number of moles of DDT formed x molar mass = 5.07 mol x 354.49 g/mol = 1799.64 g

Therefore, the mass of DDT formed is 1799.64 g.

B) To determine the limiting reactant and the reactant in excess, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation.

Using the stoichiometric ratio, the mole ratio between chlorobenzene and chloral is 2:1.

Moles of chloral needed = moles of chlorobenzene / 2 = 10.14 mol / 2 = 5.07 mol

Since we have 4.06 mol of chloral, which is less than the required 5.07 mol, it is the limiting reactant. Chlorobenzene is in excess.

C) The excess reactant is chlorobenzene.

To find the mass of the excess reactant left over, we need to calculate the moles of chlorobenzene used and subtract it from the initial moles of chlorobenzene.

Moles of chlorobenzene used = moles of chloral * 2 = 4.06 mol * 2 = 8.12 mol

Moles of chlorobenzene left over = moles of chlorobenzene initially - moles of chlorobenzene used = 10.14 mol - 8.12 mol = 2.02 mol

Mass of the excess reactant left over = moles of chlorobenzene left over * molar mass of chlorobenzene = 2.02 mol * 112.56 g/mol = 227.1312 g (rounded to four significant figures)

Therefore, the mass of the excess reactant (chlorobenzene) left over is 227.1 g.

D) The percent yield is calculated by dividing the actual yield by the theoretical yield (calculated mass of DDT) and multiplying by 100.

Percent yield = (Actual yield / Theoretical yield) x 100
Percent yield = (200.0 g / 1799.64 g) x 100 = 11.12% (rounded to two significant figures)

Therefore, the percent yield is 11.12%.

To solve the given problem, we can use stoichiometry and the concept of limiting reagents to determine the mass of DDT produced, identify the limiting reagent, calculate the remaining mass of the excess reagent, and determine the percent yield.

A) To find the mass of DDT formed, we need to determine the limiting reagent, which is the reactant that is completely consumed first. The balanced equation tells us that 2 moles of chlorobenzene react with 1 mole of chloral to produce 1 mole of DDT.

First, convert the given masses of chlorobenzene and chloral to moles using their molar masses:

Molar mass of chlorobenzene (C6H5Cl) = 112.57 g/mol
Moles of chlorobenzene = mass / molar mass = 1142 g / 112.57 g/mol ≈ 10.14 mol

Molar mass of chloral (C2HOCl3) = 147.39 g/mol
Moles of chloral = mass / molar mass = 485 g / 147.39 g/mol ≈ 3.29 mol

From the balanced equation, we can see that 2 moles of chlorobenzene react with 1 mole of DDT. Therefore, the moles of DDT formed will be half of the moles of chlorobenzene used.

Moles of DDT = 1/2 * moles of chlorobenzene = 1/2 * 10.14 mol ≈ 5.07 mol

Now, calculate the mass of DDT formed using its molar mass:

Molar mass of DDT (C14H9Cl5) = 354.48 g/mol
Mass of DDT = moles of DDT * molar mass = 5.07 mol * 354.48 g/mol ≈ 1796 g

Therefore, the mass of DDT formed is approximately 1796 grams.

B) To determine which reactant is the limiting reagent, compare the mole ratio of the reactants in the balanced equation. The reaction requires 2 moles of chlorobenzene for every 1 mole of chloral.

From the calculations above, we have:
Moles of chlorobenzene = 10.14 mol
Moles of chloral = 3.29 mol

Since the mole ratio between chlorobenzene and chloral is 2:1, and we have more than twice the amount of chlorobenzene than chloral (10.14 mol > 2 * 3.29 mol), the limiting reagent is chloral.

The reactant in excess is chlorobenzene.

C) To determine the mass of the excess reactant left over, we first need to calculate the moles of the limiting reactant that reacted.

Since the limiting reagent is chloral, and the mole ratio between chloral and DDT is 1:1, the moles of chloral reacted is equal to the moles of DDT formed.

Moles of chloral reacted = moles of DDT = 5.07 mol

Now, we can find the moles of chlorobenzene left over:

Moles of chlorobenzene left over = moles of chlorobenzene - moles of chloral reacted
Moles of chlorobenzene left over = 10.14 mol - 5.07 mol = 5.07 mol

Finally, calculate the mass of the excess chlorobenzene using its molar mass:

Mass of chlorobenzene left over = moles of chlorobenzene left over * molar mass of chlorobenzene
Mass of chlorobenzene left over = 5.07 mol * 112.57 g/mol ≈ 570 g

Therefore, the mass of the excess chlorobenzene left over is approximately 570 grams.

D) To calculate the percent yield, we need to compare the actual yield (given as 200.0 g) to the theoretical yield obtained from the limiting reagent (1796 g).

Percent yield = (actual yield / theoretical yield) * 100
Percent yield = (200.0 g / 1796 g) * 100 ≈ 11.14%

Therefore, the percent yield is approximately 11.14%.