Answer:
Theoretical yield = 59.2 g
Percent yield = 85 %
Explanation:
This is the reaction:
2Al + 3Br₂ → 2AlBr₃
If the bromine is in excess, we must think that Al is the limiting reagent.
6 g/ 26.98 g/m = 0.222 moles
As ratio is 2:2, the same amount of reactant will produce the same amount of product. Then, 0.222 moles of AlBr₃ are produced.
Molar mass . Mol = Mass
266.68 g/m . 0.222 m = 59.2 g (This is the theoretical yield)
The percent yield will be:
(Produced yield / Theoretical yield) . 100
(50.3 g / 59.2 g) . 100 = 85%
Answer:
The theoretical yield is 59.2 gram of AlBr3
The % yield is 85.0 %
Explanation:
Step 1: Data given
Mass of aluminium = 6.0 grams
bromine is in excess
Mass of aluminium bromide = 50.3 gram yield
Molar mass of Al = 26.98 g/mol
Molar mass of Br2 = 159.8 g/mol
Molar mass of AlBr3 = 266.69 g/mol
Step 2: The balanced equation
2Al + 3Br2 → 2AlBr3
Step 3: Calculate moles aluminium
moles Al = mass Al / molar mass Al
moles Al = 6.0 grams / 26.98 g/mol
moles Al = 0.222 moles
Step 4: Calculate moles AlBr3
For 2 moles Al we need 3 moles Br2 to produce 2 moles AlBr3
For 0.222 moles Al we'll have 0.222 moles AlBr3
Step 5: Calculate mass AlBr3
Mass AlBr3 = 0.222 moles * 266.69 g/mol
Mass AlBr3 = 59.2 grams
Step 6: Calculate % yield
% yield = (actual yield / theoretical yield )*100%
% yield = (50.3 grams / 59.2 grams)*100%
% yield = 85.0 %
The theoretical yield is 59.2 gram of AlBr3
The % yield is 85.0 %
