Question

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

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.

Answer 1

A.) Balanced equation for the reaction:

`2Al + 3Cl_(2)` →
`2AlCl_(3)`

5.3 mol Al ×
`(2molAlCl_(3) )/(2molAl)` ×
`(133.34g AlCl_(3) )/(1moleAlCl_(3) )` = 706.7g of
`AlCl_(3)` is produced as product

3 mol Cl ×
`(2molAlCl_(3) )/(3molCl)` ×
`(133.34gAlCl_(3) )/(1moleAlCl_(3) )` = 266.68g of
`AlCl_(3)` is produced as product

B.) The reactant that produces the list amount of product is the limiting reactant. Hence, Cl2 is limiting, while Al is in excess.

C.) moles of product formed is 2 moles.

D.) No of moles of the remaining excess reactant:

1 mole of CL - 3/2 mol of Al

3 mol of Cl - 3 ×
`(3)/(2)` = 4.5 moles of Al.

5.3 - 4.5 mole = 0.8 mol of Al remains.