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Aluminum reacts with excess copper (II) sulfate according to the reaction given below. If 1.85 g of Al react and the percentage yield of Cu is 56.6%, what mass of Cu is produced?

User Shcherbak
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Final answer:

To find the mass of Cu produced from the reaction of 1.85 g of Al with excess copper (II) sulfate at a yield of 56.6%, calculate the moles of Al, determine the theoretical yield of Cu using stoichiometry, and apply the percentage yield to find the actual yield.

Step-by-step explanation:

To calculate the mass of Cu produced when 1.85 g of Al reacts with excess copper (II) sulfate with a percentage yield of 56.6%, we need to follow these steps:

Unfortunately, the balanced chemical equation is not provided, but assuming it is similar to the copper sulfate and zinc reaction where the stoichiometry would be 1 mol Al to 1 mol Cu, we can proceed with this assumption. The molar mass of Al is 26.98 g/mol, which we use to find the moles of Al:

Moles of Al = 1.85 g / 26.98 g/mol

Next is to find the theoretical yield of Cu:

Theoretical yield of Cu = Moles of Al × (1 mol Cu / 1 mol Al) × 63.55 g/mol (Molar mass of Cu)

Finally, we calculate the actual yield using the percentage yield:

Actual yield of Cu = Theoretical yield of Cu × (56.6% / 100%)

Carry out these calculations, and you will obtain the mass of Cu produced. (Note: Be sure to use the correct balanced equation if it differs from the one assumed here.)

User Higuita
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