Respuesta :
Answer:
52.6 g AlCl3
Explanation:
266.4 g AlCl3 ->480 g Br2
29. 2 g AlCl3 -> x
x= (29.2 g AlCl3 * 480 g Br2)/266.4 g AlCl3 x= 52.6 g AlCl3
52.6 grams of Br₂ are required to react completely with 29.2 grams of AlCl₃.
What is Stoichiometry ?
Stoichiometry helps us use the balanced chemical equation to measures quantitative relationships and it is to calculate the amount of products and reactants that are given in a reaction.
The given equation is
2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂
Molar mass of AlCl₃ = Atomic weight of Al + 3 (Atomic weight of Cl)
= 27 + 3 (35.5)
= 133.5 g/mol
Molar Mass of Br₂ = 2 (Atomic weight of Br)
= 2 (79.9)
= 159.8 g/mol
Now,
[tex]29.2 \times \frac{1\ \text{mol}\ AlCl_3}{133.5\ g\ AlCl_3} \times \frac{3\ \text{mol}\ Br_2}{2\ \text{mol}\ AlCl_3} \times \frac{159.8\ g\ Br_2}{1\ \text{mol}\ Br_2}[/tex]
= 52.6 grams
Thus, we can say that 52.6 grams of Br₂ are required to react completely with 29.2 grams of AlCl₃.
Hence, Option (2) is correct answer
Learn more about the Stoichiometry here: https://brainly.com/question/14935523
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