Aluminum reacts with chlorine gas to form aluminum chloride via the formula pattern: 2Al + 3Cl2 ➝ 2AlCl3a. Identify the limiting reagentb. How many grams of aluminum chloride could be produced from 34.0g of aluminum and 39.0g of chlorine gas?c. What mass of excess reagent remains in the reaction?

Question
Asked Oct 28, 2019

Aluminum reacts with chlorine gas to form aluminum chloride via the formula pattern: 2Al + 3Cl2 ➝ 2AlCl3

a. Identify the limiting reagent

b. How many grams of aluminum chloride could be produced from 34.0g of aluminum and 39.0g of chlorine gas?

c. What mass of excess reagent remains in the reaction?

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Expert Answer

Step 1

The limiting reactant is determined as,

Al 34.0 g/ 26.98 g/mol 1.2602 mol
Cl2 39.0 g/ 70.906 g/mol 0.5500 mol
Al 1.2602 mol /2
Cl2 0.5500 mol/3
Chlorine is the limiting reactant
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Al 34.0 g/ 26.98 g/mol 1.2602 mol Cl2 39.0 g/ 70.906 g/mol 0.5500 mol Al 1.2602 mol /2 Cl2 0.5500 mol/3 Chlorine is the limiting reactant

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Step 2
Use Cl2 AICI3 molar ratio:
3 is to 2 as 0.5500 mol is to x
x 0.3667 mol of AlCl3 produced
Convert to grams:
0.3667 mol times 133.341 g/mol = 48.9 g (to three sig fig)
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Use Cl2 AICI3 molar ratio: 3 is to 2 as 0.5500 mol is to x x 0.3667 mol of AlCl3 produced Convert to grams: 0.3667 mol times 133.341 g/mol = 48.9 g (to three sig fig)

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