If we react 12 grams of aluminum with 12 grams of bromine, what is the mass of aluminum bromide?

Answer 1

Abby,

The first step to answering this question is writing the balanced equation of reaction:

Al (s) + Br₂ (l) ----> AlBr₃ (s) Unbalanced

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

Now we can determine the quantity of aluminum bromide produced, AND PICK THE SMALLEST MASS

g Al ----> mol Al-----> mol AlBr₃------>g AlBr₃

12 g * 1molAl / 26.98 g * 2 mol AlBr₃ /2 mol Al * 266.7 g/ 1 mol AlBr₃

= 119. g AlBr₃

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g Br₂ ----> mol Br₂----> mol AlBr₃------>g AlBr₃

12 g * 1mol Br₂ / 159.6 g * 3 mol AlBr₃ /2 mol Al * 266.7 g/ 1 mol AlBr₃

= 15.05 g AlBr₃

Hence the mass of aluminum yielded is 15.05 g.

Disclaimer: I did not consider sig figs and high accuracy but what I hope is that you get the method.

Answer 2

The balanced reaction follows:

2Al + 3Br₂ → 2AlBr₃

Next, we need to determine the limiting reagent:

For Al:

(12 g Al)(1 mol Al / 26.98 g Al)(2 mol AlBr₃ / 2 mol Al) = 0.44 mol AlBr₃

For Br₂:

(12 g Br₂)(1 mol Br₂ / 159.8 g Br₂)(2 mol AlBr₃ / 3 mol Br₂) = 0.05 mol AlBr₃

Therefore, Br₂ is the limiting reagent because less AlBr₃ is produced given the initial values.

Now we solve for the amount of AlBr₃ produced:

(12 g Br₂)(1 mol Br₂ / 159.8 g Br₂)(2 mol AlBr₃ / 3 mol Br₂)(266.69 g AlBr₃ / 1 mol AlBr₃) = 13.35 g AlBr₃