Final answer:
You need to convert 29.2 grams of AlCl₃ to moles, use the stoichiometric ratio from the equation to find the moles of Br₂ needed, then convert those moles to grams. The calculation shows that 52.6 grams of Br₂ are required.
Step-by-step explanation:
To determine how many grams of Br₂ are required to react completely with 29.2 grams of AlCl₃, we first need to convert the given mass of AlCl₃ to moles using its molar mass. The molar mass of AlCl₃ is 133.33 g/mol. By dividing 29.2 grams of AlCl₃ by the molar mass, we find the moles of AlCl₃.
Next, we can use the stoichiometry of the balanced equation, which shows that 2 moles of AlCl₃ react with 3 moles of Br₂. Using this ratio, we can determine the moles of Br₂ needed. Lastly, we convert the moles of Br₂ to grams using its molar mass, which is 159.808 g/mol (79.904 g/mol for each Br atom times two).
Here's the step-by-step calculation:
- Calculate moles of AlCl₃: 29.2 g / 133.33 g/mol = 0.219 moles of AlCl₃
- Use the balanced equation to calculate moles of Br₂ needed: (3 moles of Br₂ / 2 moles of AlCl₃) × 0.219 moles of AlCl₃ = 0.3285 moles of Br₂
- Convert moles of Br₂ to grams: 0.3285 moles × 159.808 g/mol = 52.5 grams of Br₂ (rounded to one decimal place)
Therefore, 52.6 grams of Br₂ are required, which corresponds to choice b).