If 3.0 g of aluminum and 6.0 g of bromine react to form AlBr3, how many grams of product would theoretically be produced? How many grams of each reagent would remain at the end of this reaction?

What mass (in grams) of product would be collected if the reaction above proceeded in 72% yield?

First, we determine the moles of aluminum and bromine present using their atomic masses and given masses:

moles = mass / atomic mass

Aluminum:

3 / 27 = 0.11 mole

Bromine:

6 / 80 = 0.075

The molar ratio of bromine to aluminum in the product is 3 : 1, so the moles of aluminum that will be needed are:

0.075 / 3 = 0.025

So, 0.025 moles of product will be formed as the reaction is limited due to bromine. This means the mass of product will be:

Mass = moles * Mr

Mass = 0.025 * 207

Mass = 5.18 grams of product

The remaining reagents will be:

Br - 0 grams as all will react

Al - (0.11 - 0.025) * 27 = 2.30 grams as it is in excess

If the reaction was to go to 72% yield, the mass formed would be 72% of the mass formed when the reaction completes. So:

Mass = 0.72 * 5.18

Mass = 3.73 grams