Answer:
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.
Step-by-step explanation:
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= 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 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.