View Available Hint(s) Check all that apply. Hydrogen bonding occurs when a hydrogen atom is covalently bonded to an NN, OO, or
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Answer : The correct answer for the Theoretical Yield is 48.93 g of product .
Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .
Following are the steps to find theoretical yield .
Step 1) : Write a balanced reaction between Al and Cl₂ .
2 Al + 3 Cl₂→ 2 AlCl₃
Step 2: To find amount of product (AlCl₃) formed by Al .
Following are the sub steps to calculate amount of AlCl₃ formed :
a) To calculate mole of Al :
Given : Mass of Al = 35.5 g
Mole can be calculate by following formula :
Mole = 1.32 mol
b) To find mole ratio of AlCl₃ : Al
Mole ratio is calculated from balanced reaction .
Mole of Al in balanced reaction = 2
Mole of AlCl₃ in balanced reaction = 2.
Hence mole ratio of AlC; l₃ : Al = 2:2
c) To find mole of AlCl₃ formed :
Mole of AlCl₃ = 1.32 mol
d) To find mass of AlCl₃
Molar mass of AlCl₃ =
Mass of AlCl3 can be calculated using mole formula as:
Multiplying both side by
Mass of AlCl₃ = 176.00 g
Hence mass of AlCl₃ produced by Al is 176.00 g
Step 3) To find mass of product (AlCl₃) formed by Cl₂ :
Same steps will be followed to calculate mass of AlCl₃
a) Find mole of Cl₂
Mole of Cl₂ = 0.55 mol
b) Mole ratio of Cl₂ : AlCl₃
Mole of Cl₂ in balanced reaction = 3
Mole of AlCl₃ in balanced reaction = 2
Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3
c) To find mole of AlCl₃
Mole of AlCl3 = 0.367 mol
d) To find mass of AlCl₃ :
Multiplying both side by
Mass of AlCl₃ = 48.93 g
Hence mass of AlCl₃ produced by Cl₂ = 48.93 g
Step 4) To identify limiting reagent and theoretical yield :
Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .
The product AlCl₃ formed by Al = 176.00 g
The product AlCl₃ formed by Cl₂ = 48.93 g
Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .
Mass of Oxygen: 0.0159 grams
Moles of Oxygen: 9.94x10^-4
To find the mass of oxygen, subtract the mass of copper from the total mass.
There are 0.0159 grams of Oxygen.
To find how many moles there are, divide the given amount of oxygen by the molar mass (atomic mass) of oxygen because that mass is the same as one mole of oxygen.
Molar mass of Oxygen: 16.00
There are 9.94*10^-4 moles of Oxygen.