Question

1) ____Al + ____ Cl₂ _____ AlCl₃If you have 1.50 mole of Aluminum and excess chlorine, what mass of AlCl₃ can you produce?If you made 10.8 moles of AlCl₃, how many moles of Aluminum would you need?If 25.0 moles of chlorine is used, how many grams of aluminum chloride can be produced?

Answer 1

Step 1

The reaction must be written and balanced:

2Al + 3Cl₂ 2AlCl₃

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Step 2

Information provided:

1.50 moles of Al (the limiting reactant)

Information needed: the molar mass of AlCl3 => 133 g/mol

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Step 3

Procedure:

2Al + 3Cl₂ 2AlCl₃ by stoichiometry:

2 moles Al -------- 2 x 133 g AlCl3

1.50 moles Al -------- X

X = 1.50 moles Al x 2 x 133 g AlCl3/2 moles Al = 199.5 moles

Answer: 199.5 moles AlCl3