Answer:
Explanation:
A. How many kilograms of C₁₁H₇O₂Cl form from 2.5×10*2 kg of naphthol?
Given data:
Mass of naphthol = 2.5 ×10² kg ( 250×1000 = 250000 g)
Mass of C₁₁H₇O₂Cl = ?
Solution:
Chemical equation:
C₁₀H₈O + COCl₂ → C₁₁H₇O₂Cl + HCl
Number of moles of naphthol :
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 data:
Mass of naphthol = 100 g
Mass of COCl₂ = 100 g
Theoretical yield of C₁₁H₇O₂Cl = ?
Solution:
Chemical equation:
C₁₀H₈O + COCl₂ → C₁₁H₇O₂Cl + HCl
Number of moles of naphthol:
Number of moles = mass/ molar mass
Number of moles = 100 g/ 144.17 g/mol
Number of moles = 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 data:
Actual yield of C₁₁H₇O₂Cl = 118 g
Percent yield = ?
Solution:
Formula :
Percent yield = actual yield / theoretical yield × 100
Theoretical yield = 143.3 g (from part b)
Now we will put the values in formula.
Percent yield = 118 g/ 143.3 g × 100
Percent yield = 0.82 × 100
Percent yield = 82%
