Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Explore in-depth answers to your questions from a knowledgeable community of experts across different fields. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
Answer:
296.67 g of AlCl₃.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2Al + 6HCl —>2AlCl₃ + 3H₂
Next, we shall determine the mass of Al that reacted and the mass of AlCl₃ produced from the balanced equation. This can be obtained as follow:
Molar mass of Al = 27 g/mol
Mass of Al from the balanced equation = 2 × 27 = 54 g
Molar mass of AlCl₃ = 27 + (35.5×3)
= 27 + 106.5
= 133.5 g/mol
Mass of AlCl₃ from the balanced equation = 2 × 133.5 = 267 g
SUMMARY:
From the balanced equation above,
54 g of Al reacted to produce 267 g of AlCl₃.
Finally, we shall determine the mass of AlCl₃ produced by the reaction of 60 g of Al. This can be obtained as follow:
From the balanced equation above,
54 g of Al reacted to produce 267 g of AlCl₃.
Therefore, 60 g of Al will react to produce = (60 × 267)/54 = 296.67 g of AlCl₃
Thus, 296.67 g of AlCl₃ were obtained from the reaction.
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.