In one step in the synthesis of the insecticide Sevin, naphthol reacts with phosgene as shown.

C10H8O+COCl2⟶C11H7O2Cl+HCl

Naphthol Phosgene

A. How many kilograms of C11H7O2Cl form from 2.5*10*2 kg of naphthol?

B. If 100. g of naphthol and 100. g of phosgene react, what is the theoretical yield of C11H7O2Cl?

C. If the actual yield of C11H7O2Cl in part b is 118 g, what is the percent yield?

Chemical equation:

C₁₀H₈O + COCl₂ → C₁₁H₇O₂Cl + HCl

A. How many kilograms of C₁₁H₇O₂Cl form from 2.5*10*2 kg of naphthol?

Given dа ta:

Mass of naphthol = 2.5 * 10² kg (250*1000 = 250000 g)

Mass of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol = mass / molar mass

Number of moles of naphthol = 250000 g / 144.17 g/mol

Number of moles of naphthol = 1734.1 mol

Now we will compare the moles of naphthol with C₁₁H₇O₂Cl.

C₁₀H₈O : C₁₁H₇O₂Cl

1 : 1

1734.1 : 1734.1

Mass of C₁₁H₇O₂Cl:

Mass = number of moles * molar mass

Mass = 1734.1 mol * 206.5 g/mol

Mass = 358091.65 g

Gram to kilogram:

1 kg*358091.65 g / 1000 g = 358.1 kg

B. If 100. g of naphthol and 100. g of phosgene react, what is the theoretical yield of C11H7O2Cl?

Given dа ta:

Mass of naphthol = 100 g

Mass of COCl₂ = 100 g

Theoretical yield of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol:

Number of moles of naphthol = mass / molar mass

Number of moles of naphthol = 100 g / 144.17 g/mol

Number of moles of naphthol = 0.694 mol

Number of moles of phosgene:

Number of moles = mass / molar mass

Number of moles = 100 g / 99 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of naphthol and phosgene with C₁₁H₇O₂Cl.

C₁₀H₈O : C₁₁H₇O₂Cl

1 : 1

0.694 : 0.694

COCl₂ : C₁₁H₇O₂Cl

1 : 1

1.0 : 1.0

The number of moles of C₁₁H₇O₂Cl produced by C₁₀H₈O are less so it will limiting reactant and limit the yield of C₁₁H₇O₂Cl.

Mass of C₁₁H₇O₂Cl:

Mass = number of moles * molar mass

Mass = 0.694 mol * 206.5 g/mol

Mass = 143.3 g

Theoretical yield = 143.3 g

C. If the actual yield of C11H7O2Cl in part b is 118 g, what is the percent yield?

Given dа ta:

Actual yield of C₁₁H₇O₂Cl = 118 g

Theoretical yield = 143.3 g

Percent yield = ?

Solution:

Formula:

Percent yield = actual yield / theoretical yield * 100

Now we will put the values in formula.

Percent yield = 118 g / 143.3 g * 100

Percent yield = 0.82 * 100

Percent yield = 82%