Aluminum burns in bromine producing aluminum bromide in a certain experiment, 6.0 g aluminum was reacted with an excess of bromine to yield 50.3 g aluminum bromide calculate the theoreteical and percent yields.

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%

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 %

it's filtration, but this is an assumption so my apologies if i got this