Question

Aluminum chloride, AlCl3, is an inexpensive reagent used in many industrial processes. It is made by treating scrap aluminum with chlorine according to the following equation. Please balance the equation.
__2_ Al (s) + 3___ Cl2(g) _2___ AlCl3(s)
If you start with 3.11 g of Al and 5.32 g of Cl2, which reagent is limiting? How many grams of AlCl3 can be produced? During an experiment you obtained 5.57 g of AlCl3, what was your percent yield?

Answer 1

Answer:

83.8%

Explanation:

The balanced reaction equation is;

2Al(s) + 3Cl2(g) → 2AlCl3(s)

Now we have to obtain the limiting reactant as the reactant that produces the least amount of AlCl3

Amount of Al = 3.11g/27 g/mol = 0.115 moles

If 2 moles of Al yields 2 moles of  AlCl3

Then 0.115 moles of Al yields 0.115 moles of  AlCl3

For Cl2

Amount of Cl2 = 5.32 g/71 g/mol= 0.075 moles

If 3 moles of Cl2 yields 2 moles of  AlCl3

0.075 moles of Cl2 yields 0.075   * 2/3 = 0.05 moles of  AlCl3

Hence Cl2 is the limiting reactant

Theoretical yield of  AlCl3 = 0.05 moles of  AlCl3 * 133g/mol = 6.65 g

%yield = actual yield /theoretical yield * 100

%yield = 5.57 g/6.65 g * 100

%yield = 83.8%