Question

3.0 g aluminum and 6.0 g of bromine react to form AlBr3

2Al+3Br2=2AlBr3

How much product would be produced?

How much reagent would remain at the end of this reaction?

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

Answer 1

Hint: Law of conservation mass states that whatever you have at the beginning will stay throughout the chemical reaction till the end. You can use dimensional analysis to calculate this

Answer 2

2Al + 3Br2 ==> 2AlBr3
m Al = 3 grams
m Br = 6 grams

m AlBr = ?

First, you need to find moles for Al and Br using the masses provided.
n = m / M
You can find molar mass (M) using your periodic table.

After you find moles, you must do a mole-mole ratio to find the number of moles in your product, AlBr3.

# moles Al (you got from previous step) x 2 moles AlBr3 / 2 moles Al

The moles for Al are cancelled, or divided out, and you are left with moles of AlBr3.
Using this mole value, you can find mass.

You can do this using the Br as well. I don't think it matters.