Question
Asked Nov 8, 2019
Aluminum chloride, AlCl3, is made by treating scrap aluminum with chlorine.

2 Al(s) + 3 Cl2(g) → 2 AlCl3(s)

If you begin with 7.80 g Al and 3.11 g Cl2, identify the limiting reactant and determine what mass of the excess reactant remains when the reaction is completed?
 

 

 

check_circleExpert Solution
Step 1

Given,

Mass of Al = 7.80 g

Mass of Cl2 = 3.11 g

Moles of Al and Cl2 can be calculated as:

Molar mass of Al = 27 g/mol
Molar mass of Cl2 (2x35.5) = 71 g/mol
7.80 g
Mass
Moles of Al
0.289 mol
Molar mass
27 g/mol
3.11 g
Mass
Moles of Cl2=
0.0438 mol
Molar mass
71 g/mol
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Molar mass of Al = 27 g/mol Molar mass of Cl2 (2x35.5) = 71 g/mol 7.80 g Mass Moles of Al 0.289 mol Molar mass 27 g/mol 3.11 g Mass Moles of Cl2= 0.0438 mol Molar mass 71 g/mol

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Step 2

The balanced reaction of scrap aluminium with chlorine is given as:

2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s)

From the balanced reaction:

2 mol Al requires
3 mol Cl2
3 mol Cl2
.0.289 mol Al requires
x 0.289 mol Al
2 mol Al
0.4335 mol Cl2
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2 mol Al requires 3 mol Cl2 3 mol Cl2 .0.289 mol Al requires x 0.289 mol Al 2 mol Al 0.4335 mol Cl2

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Step 3

Since, 0.4335 mol of Cl2 is required which is less than the actual moles of Cl2 (0.0438 mol) There...

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