Question

Find the mass of AlCl3 that is produced when 10.0 grams of Al2O3 react with 10.0 g of HCl according to the following equation. Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(aq)

Answer 1

Answer: The mass of aluminium chloride produced is 12.17 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

• For aluminium oxide:

Given mass of aluminium oxide = 10.0 g

Molar mass of aluminium oxide = 102 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium oxide}=\frac{10g}{102g/mol}=0.098mol$

• For HCl:

Given mass of HCl = 10 g

Molar mass of HCl = 36.5 g/mol

Putting values in equation 1, we get:

$\text{Moles of HCl}=\frac{10g}{36.5g/mol}=0.274mol$

The given chemical equation follows:

$Al_2O_3(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2O(aq.)$

By Stoichiometry of the reaction:

6 moles of HCl reacts with 1 mole of aluminium oxide

So, 0.274 moles of HCl will react with = $\frac{1}{6}\times 0.274=0.046mol$ of aluminium oxide

As, given amount of aluminium oxide is more than the required amount. So, it is considered as an excess reagent.

Thus, HCl is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

6 moles of HCl produces 2 moles of aluminium chloride

So, 0.274 moles of HCl will produce = $\frac{2}{6}\times 0.274=0.0913moles$ of aluminium chloride

Now, calculating the mass of aluminium chloride from equation 1, we get:

Molar mass of aluminium chloride = 133.34 g/mol

Moles of aluminium chloride = 0.0913 moles

Putting values in equation 1, we get:

$0.0913mol=\frac{\text{Mass of aluminium chloride}}{133.34g/mol}\\\\\text{Mass of aluminium chloride}=(0.0913mol\times 133.34g/mol)=12.17g$

Hence, the mass of aluminium chloride produced is 12.17 grams

Answer 2

Considering the reaction stoichiometry and the definition of limiting reagent, the mass of AlCl₃ that is produced when 10.0 grams of Al₂O₃ react with 10.0 grams of HCl is 12.19 grams.

The balanced reaction is:

Al₂O₃ + 6 HCl → 2 AlCl₃ + 3 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:  

• Al₂O₃: 1 mole
• HCl: 6 moles
• AlCl₃: 2 moles
• H₂O: 3 moles

The molar mass of the compounds present in the reaction is:

• Al₂O₃: 102 g/mole
• HCl: 36.45 g/mole
• AlCl₃: 133.35 g/mole
• H₂O: 18 g/mole

Then, by reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of mass of each compound participate in the reaction:  

• Al₂O₃: 1 mole× 102 g/mole= 102 grams
• HCl: 6 moles× 36.45 g/mole= 218.7 grams
• AlCl₃: 2 moles× 133.35 g/mole= 266.7 grams
• H₂O: 3 moles× 18 g/mole= 54 grams

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 102 grams of Al₂O₃ reacts with 218.7 grams of HCl, 10 grams of Al₂O₃ reacts with how much moles of HCl?

$mass of HCl=\frac{10 grams of Al_{2} O_{3} x218.7 grams of HCl}{102 grams of Al_{2} O_{3}}$

mass of HCl= 21.44 grams

But 21.44 grams of HCl are not available, 10 grams are available. Since you have less moles than you need to react with 10 grams of Al₂O₃ , HCl will be the limiting reagent.

Then, it is possible to determine the mass of AlCl₃ produced by another rule of three: if by stoichiometry 218.7 grams of HCl produce 266.7 grams of AlCl₃, if 10 grams of HCl react how much mass of AlCl₃ will be formed?

$mass of AlCl_{3} =\frac{10 grams of HClx266.7 grams of AlCl_{3} }{218.7 grams of HCl}$

mass of AlCl₃= 12.19 grams

In summary, the mass of AlCl₃ that is produced when 10.0 grams of Al₂O₃ react with 10.0 grams of HCl is 12.19 grams.

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