Producing Aluminum Chloride

Problem

Aluminum chloride (AlCl₃) is produced by reacting aluminum and chlorine gas.

Part 1: Determine the chemical formula.

Part 2: What is the limiting reagent if 10.0 grams of aluminum react with 15.0 grams of chlorine.

Part 3: Determine the mass of AlCl₃ that could be produced from part 2. This mass is called the theoretical yield.

Solution

Part 1

Chlorine gas is diatomic, so begin with the guess:

$Al + Cl_2 \rightarrow AlCl_3$

Balance the equation.

$2Al + 3Cl_2 \rightarrow 2AlCl_3$

Part 2

Perform a stoichiometric calculation with the given information for aluminum.

Failed to parse (syntax error): {\displaystyle 10.0 \:g \left(\frac{1 \: mol \: Al}{27.0 \: g \: Al}\right) = 0.370 \: mol \: Al }

Failed to parse (syntax error): {\displaystyle 0.370 \: mol \: Al \left(\frac{3 \: Cl_2}{2 \: Al}\right) = 0.556 \: mol \: Cl_2 }

Failed to parse (syntax error): {\displaystyle 0.556 \: mol \: Cl_2 \left(\frac{71.0 \: g \: Cl_2}{1 \: mol \: Cl_2}\right) = 39.4 \: g \: Cl_2 }

We only have 15.0 grams of chlorine, so the chlorine will run out first. This means chlorine is the limiting reagent.

Part 3

How much aluminum chloride (in grams) is produced?

Failed to parse (syntax error): {\displaystyle 15.0 \: g \: Cl_2 \left(\frac{1 \: mol \: Cl_2}{70.0 \: g \: Cl_2}\right) = 0.214 \: mol \: Cl_2 }

Failed to parse (syntax error): {\displaystyle 0.214 \: mol \: Cl_2 \left(\frac{2 \: Cl_2}{3 \: Al}\right) = 0.143 \: mol \: AlCl_3 }

Failed to parse (syntax error): {\displaystyle 0.143 \: mol \: AlCl_3 \left(\frac{133.5 \: g \: AlCl_3}{1 \: mol \: AlCl_3}\right) = 19.1 \: g \: AlCl_3 }

The theoretical yield aluminum chloride produced from the above process is 19.1 grams.