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Answer: The correct answer is about 0.20 grams of copper (II) is formed, and some aluminum is left in the reaction mixture.
Explanation:
To calculate the number of moles, we use the formula:
- <u>Moles of Aluminium:</u>
Molar mass of aluminium = 27 g/mol
Given mass of aluminium = 0.25g
Putting values in above equation, we get:
- <u>Moles of Copper chloride:</u>
Molar mass of copper chloride = 134.45 g/mol
Given mass of copper chloride= 0.40g
Putting values in above equation, we get:
For the given chemical equation:
By Stoichiometry,
3 moles of Copper chloride reacts with 4 moles of aluminium
So, 0.00297 moles of copper chloride will react with = = 0.00396 moles of Aluminium.
As, the required moles of aluminium is less than the given moles of aluminium. Hence, it is considered as the excess reagent.
Copper chloride is the limiting reagent in the given chemical reaction because it limits the formation of products.
By Stoichiometry of the reaction:
3 moles of copper chloride produces 3 moles of copper metal.
So, 0.00297 moles of copper chloride will produce = = 0.00297 moles of copper metal.
Now, to calculate the given mass of copper metal, we use the equation required to calculate moles:
Molar mass of copper = 63.5 g/mol
Putting values in that equation, we get:
Given mas of copper metal = 0.188grams (approx 0.20grams)
For the given reaction, some of the aluminum is left behind because it is the excess reagent in the reaction.
Hence, the correct answer is about 0.20 grams of copper (II) is formed, and some aluminum is left in the reaction mixture.