Answer:
a) percent yield of CH₄ = 85.5 %
b) percent yield of AgBr = 79 %
c) percent yield of FeCl₃ = 82 %
Explanation:
Part a:
Data given:
mass of carbon (C) = 1.4 g
actual yield of methane (CH₄) = 1.6 g
percent yield of methane (CH₄) = ?
Reaction Given:
C + 2H₂ -----> CH₄
Solution:
First we have to find theoretical yield.
So,
Look at the reaction
C + 2H₂ -----> CH₄
1 mol 1 mol
As 1 mole of C give 1 mole of CH₄
Convert moles to mass
molar mass of C = 12 = 12 g/mol
molar mass of CH₄ = 12 + 4(1) = 16 g/mol
Now
C + 2H₂ -------> CH₄
1 mol (12 g/mol) 1 mol (16 g/mol)
12 g 16 g
12 g of carbon (C) produce 16g of CH₄
So
if 12 g of carbon (C) produce 16g of CH₄ so how many grams of CH₄ will be produced by 1.4 g of C.
Apply Unity Formula
12 g of carbon (C) ≅ 16g of CH₄
1.4 g of carbon (C) ≅ X g of CH₄
Do cross multiply
mass of CH₄ = 16 g x 1.4 g / 20 g
mass of CH₄ = 1.87 g
So the Theoretical yield of CH₄ = 1.87 g
Now Find the percent yield of CH₄
Formula Used
percent yield = actual yield /theoretical yield x 100 %
Put value in the above formula
percent yield = 1.6 g / 1.87 g x 100 %
percent yield = 85.5 %
percent yield of CH₄ = 85.5 %
______________________________
Part b:
Data given:
mass of Br₂ = 3.8 g
actual yield of AgBr = 6.7 g
percent yield of AgBr = ?
Reaction Given:
2Ag + Br₂ -----> 2AgBr
Solution:
First we have to find theoretical yield.
So,
Look at the reaction
2Ag + Br₂ -----> 2AgBr
1 mol 2 mol
As 1 mole of Br₂ give 2 mole of AgBr
Convert moles to mass
molar mass of Br₂ = 2(80) = 160 g/mol
molar mass of AgBr = 108 + 80 = 188 g/mol
Now
2Ag + Br₂ ---------> 2AgBr
1 mol (160 g/mol) 2 mol (188 g/mol)
160 g 376 g
169 g of Br₂ produce 376 g of AgBr
So
if 169 g of Br₂ produce 376 g of AgBr then how many grams of AgBr will be produced by 3.8 g of Br₂.
Apply Unity Formula
169 g of Br₂ ≅ 376 g of AgBr
3.8 g of Br₂ ≅ X g of AgBr
Do cross multiply
mass of AgBr = 376 g x 3.8 g / 169 g
mass of AgBr = 8.45 g
So the Theoretical yield of AgBr = 8.45 g
Now Find the percent yield of AgBr
Formula Used
percent yield = actual yield /theoretical yield x 100 %
Put value in the above formula
percent yield = 6.7 g / 8.45 g x 100 %
percent yield = 79 %
percent yield of AgBr = 79 %
______________________________
Part c:
Data given:
mass of Cl₂ = 13 g
actual yield of FeCl₃ = 16.3 g
percent yield of FeCl₃ = ?
Reaction Given:
2Fe + 3Cl₂ -----> 2FeCl₃
Solution:
First we have to find theoretical yield.
So,
Look at the reaction
2Fe + 3Cl₂ -----> 2FeCl₃
3 mol 2 mol
As 3 mole of Cl₂ give 2 mole of FeCl₃
Convert moles to mass
molar mass of Cl₂ = 2(35.5) = 71 g/mol
molar mass of FeCl₃ = 56 + 3(35.5) = 162.5 g/mol
Now
2Fe + 3Cl₂ ---------> 2FeCl₃
3 mol (71 g/mol) 2 mol (162.5 g/mol)
213 g 325 g
213 g of Cl₂ produce 325 g of FeCl₃
So
213 g of Cl₂ produce 325 g of FeCl₃ then how many grams of FeCl₃ will be produced by 13 g of Cl₂.
Apply Unity Formula
213 g of Cl₂ ≅ 325 g of FeCl₃
13 g of Cl₂ ≅ X g of FeCl₃
Do cross multiply
mass of FeCl₃ = 325 g x 13 g / 213 g
mass of FeCl₃ = 19.8 g
So the Theoretical yield of FeCl₃= 19.8 g
Now Find the percent yield of FeCl₃
Formula Used
percent yield = actual yield /theoretical yield x 100 %
Put value in the above formula
percent yield = 16.3 g / 19.8 g x 100 %
percent yield = 82 %
percent yield of FeCl₃ = 82 %
