PLEASE HELP WILL MARK BRAINLIEST

5.3 mol Al reacts with 3.0 mol Cl2 to produce Aluminum chloride.

(ANSWER A, B, C and D)

a. Write and balance the chemical equation.

b. Identify the limiting reagent.

c. Calculate the moles of product formed.

d. calculate the number of moles of excess reagent remaining after the

reaction is completed.

a) 2Al + 3Cl2 ⇆ 2AlCl3

b) Cl2 is the limiting reactant.

c)  2.0 moles AlCl3

d) There will remain  3.2 moles Al

Explanation:

Step 1: Data given

Number of moles Al = 5.3 moles

Number of moles Cl2 = 3.0 moles

Step 2: The balanced equation

2Al + 3Cl2 ⇆ 2AlCl3

Step 3: Calculate the limiting reactant

For 2 moles Al we need 3 moles Cl2 to produce 2 moles AlCl3

Cl2 is the limiting reactant. It will completely be consumed. (3.0 moles)

Al is in excess. There will react 2/3 * 3.0 = 2.0 moles

There will remain 5.2 - 2.0 = 3.2 moles

Step 4: Calculate moles AlCl3

For 2 moles Al we need 3 moles Cl2 to produce 2 moles AlCl3

For 3.0 moles Cl2 we'll have 2/3 * 3.0 = 2.0 moles AlCl3

c.

explanation:

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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