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Answer:

115.36 grams of AlCl₃ will be produced from 92 grams of Cl₂.

Explanation:

The balanced reaction is:

2 AlBr₃ + 3 Cl₂ ⇒ 3 Br₂ + 2 AlCl₃

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:

  • AlBr₃: 2 moles
  • Cl₂: 3 moles  
  • Br₂: 3 moles
  • AlCl₃: 2 moles

Being the molar mass of each compound:

  • AlBr₃: 266.7 g/mole
  • Cl₂: 70.9 g/mole
  • Br₂: 159.8 g/mole
  • AlCl₃: 133.35 g/mole

By stoichiometry of the reaction the following quantities of mass participate in the reaction:

  • AlBr₃: 2 moles* 266.7 g/mole= 533.4 g
  • Cl₂: 3 moles* 70.9 g/mole= 212.7 g
  • Br₂: 3 moles* 159.8 g/mole= 479.4 g
  • AlCl₃: 2 moles* 133.35 g/mole= 266.7 g

You can apply the following rule of three: if by reaction stoichiometry 212.7 grams of Cl₂ produce 266.7 grams of AlCl₃, 92 grams of Cl₂ will produce how much mass of AlCl₃?

[tex]mass of AlCl_{3}=\frac{92 grams of Cl_{2}*266.7 grams of AlCl_{3}}{212.7grams of Cl_{2}}[/tex]

mass of AlCl₃= 115.36 grams

115.36 grams of AlCl₃ will be produced from 92 grams of Cl₂.

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