Answer:
115.36 grams of AlCl₃ will be produced from 92 grams of Cl₂.
Step-by-step 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?
mass of AlCl₃= 115.36 grams
115.36 grams of AlCl₃ will be produced from 92 grams of Cl₂.