) 5.5 moles of Aluminum react with 5.5 moles of Sulfur to produce Aluminum Sulfide. a) What is the limiting reactant? What is the excess reagent? b) How many grams of Aluminum Sulfide will be produced? c) How many grams of the excess reactant will be left over in the reac

a) S is the limiting reagent. The excess reagent is Alent is Al.

b) 2.75 moles

c) 1.8 moles

Explanation:

(a)

We begin by writing down the reaction equation;

2Al + 3S → Al₂ S₃

The limiting agent is the one that is consumed first in the chemical reactions. For every 2 moles of Al 3 moles of S are consumed to completely react the reactants to products.

Since 5.5 moles of Al and 5.5 moles of S are being reacted, S will be consumed first, hence this is the limiting reagent.

The excess reagent is Alent is Al. This is because, for the entire 5.5 moles of AL to react, it would require the following moles of S;

2 : 3

5.5 : x

2x = 5.5 * 3

X = 8.25 moles

8.25 moles of S

(b)

We will use the mole ratio to calculate the amount of Al produced. Mole ration between Al and Al2S3 is;

2 : 1

5.5: x

2x = 5.5

x = 2.75

= 2.75 moles

(c)

To find the amount of AL left unreacted, we find the amount that reacted with the limiting reagent S (that was consumed completely);

2 : 3

x : 5.5

3x = 5.5 * 2

X = 3.67 moles

3,67 moles of AL were consumed. Therefore the amount left is;

5.5 – 3.7

= 1.8 moles

the answer is barium and copper bromide