How do you do Limiting Reagent, Theoretical Yield, and Percent Yield ?

Question

how do you do this problem?

Aluminum burns bromine, producing aluminum bromide:

2Al (s) + 3Br2 (l) ---> 2AlBr3 (s)

in a certain experiment, 6.0 g aluminum was reacted with an excess of bromine to yield 50.3g aluminum bromide. Calculate the theoretical and percent yields for this experiment?

Answer 1

Since the problem states that the aluminum was reacted with "an excess of bromine", then the aluminum is by definition your limiting reagent.

Now, convert 6.0 grams of Al to moles of Al. Using the stoichiometry of the balanced equation, then convert moles of Al into moles of AlBr3. Finally convert the moles of AlBr3 into grams of AlBr3. The mass of AlBr3 you calculated is the theoretical yield.

To calculate your %yield, divide the 50.3 grams AlBr3 obtained by the theoretical yield, and multiply that by 100. That will give you the % yield.

Answer 2

First you have to find out how many moles of Al you have.

6.0g / 27.0g/mole = 0.222 moles

You are told that there is excess bromine, so the aluminium is the limiting reagent. This is because adding more bromine to the system will not increase your yield, but adding more aluminium will.

It can be seen from the equation that every mole of Al which reacts produces 1 mole of AlBr3

In theory, you should produce 0.222 moles of AlBr3

0.222 moles of AlBr3 at 267 g/mole = 59.3g

You have 50.3g, so your % yield is 50.3/59.3) x 100 = 84.8%

Answer 3

6.0 g Al x (1 mole Al / 27.0 g Al) x (2 moles AlBr3 / 2 moles Al) x (266.7 g AlBr3 / 1 mole AlBr3) = 59.3 g AlBr3 ... this is the theoretical yield.

Percent yield = (actual yield / theoretical yield) x 100% = (50.3 g / 59.3 g) x 100% = 84.8%