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:

Cl₂ the limiting reactant.

6.67g of AlCl₃ can be produced.

83.5% is percent yield

Explanation:

Based on the reaction:

2 Al + 3Cl₂ → 2AlCl₃

2 moles of aluminium react with 3 moles of chlorine.

To solve this question we must find the moles of each reactant in order to find limiting reactant. With limiting reactant we can find theoretical yield. Percent yield is:

Actual yield (5.57g) / Theoretical yield * 100

That means if we find theoretical yield we can find percent yield:

Moles Aluminium: 26.98g/mol

3.11g * (1mol / 26.98g) = 0.115 moles Al

Moles Chlorine: 70.90g/mol

5.32g * (1mol / 70.90g) = 0.075 moles Cl₂

For a complete reaction of 0.075 moles of Cl₂ are required:

0.075 moles Cl₂ * (2mol Al / 3mol Cl₂) = 0.050 moles of Al

As there are 0.115 moles of Al, Aluminium is the excess reactant and Cl₂ the limiting reactant.

Moles AlCl₃ and mass: 133.34g/mol

0.075 moles Cl₂ * (2mol AlCl₃ / 3mol Cl₂) = 0.050 moles of AlCl₃

0.050 moles of AlCl₃ * (133.34g / mol) =

6.67g of AlCl₃ can be produced

Percent yield:

5.57g / 6.67g * 100 =

83.5% is percent yield