What is the percent yield if 4.65g of copper is produced when 1.87g of aluminum reacts with an excess of copper(II) sulfate?

2Al(s)+3CuSO4(aq)==>Al2(SO4)3(aq)+3Cu(s)

1.87 g Al = ?? moles.

moles = 1.87/molar mass Al.

Using the coefficients in the balanced equation, convert moles Al to moles Cu.

Now convert moles Cu to grams. g = moles x molar mass Cu. This is the theoretical yield of the reaction.

%yield = (actual yield/theoretical yield)*100 = ??
The actual yield is given in the problem as 4.65 grams.

To calculate the percent yield, you need to know the theoretical yield and the actual yield.

First, let's determine the theoretical yield:

1. Calculate the molar mass of aluminum (Al):
Al = 26.98 g/mol

2. Calculate the molar mass of copper (Cu):
Cu = 63.55 g/mol

3. Calculate the molar mass of copper(II) sulfate (CuSO4):
CuSO4 = (63.55 g/mol) + (32.07 g/mol) + (4 * 16.00 g/mol) = 159.61 g/mol

4. Convert the mass of aluminum (1.87 g) to moles:
moles of Al = mass / molar mass = 1.87 g / 26.98 g/mol = 0.0693 mol

5. Use the balanced equation to determine the stoichiometric ratio between aluminum and copper:
From the balanced equation, we can see that 2 moles of Al produce 3 moles of Cu.

6. Calculate the moles of copper produced:
moles of Cu = (moles of Al) × (3/2) = 0.0693 mol × (3/2) = 0.1039 mol

7. Convert moles of copper to grams:
mass of Cu = moles × molar mass = 0.1039 mol × 63.55 g/mol = 6.599 g

Now, let's calculate the percent yield:

Actual yield = 4.65 g
Theoretical yield = 6.599 g

Percent yield = (actual yield / theoretical yield) × 100
Percent yield = (4.65 g / 6.599 g) × 100 = 70.54%

Therefore, the percent yield of copper is approximately 70.54%.

To calculate the percent yield, you need to know the actual yield and the theoretical yield. The actual yield is the amount of copper actually produced in the reaction, which in this case is given as 4.65g. The theoretical yield is the maximum amount of copper that can be produced based on stoichiometry and is calculated using the balanced equation.

First, we need to find the limiting reactant to determine the amount of copper that can be produced. To do this, compare the number of moles of aluminum and copper(II) sulfate. The balanced equation shows that 2 moles of aluminum react with 3 moles of copper(II) sulfate.

To calculate the moles of aluminum, divide the given mass (1.87g) by the molar mass of aluminum (26.98 g/mol).
moles of aluminum = 1.87g / 26.98 g/mol ≈ 0.069 mol

Next, we calculate the moles of copper(II) sulfate that can react with the moles of aluminum.
moles of copper(II) sulfate = (0.069 mol Al) × (3 mol CuSO4 / 2 mol Al) = 0.1035 mol

Now that we have the moles of copper(II) sulfate that can react, we can calculate the theoretical yield of copper using the balanced equation. According to the balanced equation, the molar ratio of Cu to CuSO4 is 3:3 (3 moles of Cu for every 3 moles of CuSO4).
moles of copper = 0.1035 mol × (3 mol Cu / 3 mol CuSO4) = 0.1035 mol

Finally, we can convert the moles of copper to grams by multiplying by the molar mass of copper (63.55 g/mol).
theoretical yield of copper = 0.1035 mol × 63.55 g/mol = 6.578g

Now we can calculate the percent yield by using the formula:
percent yield = (actual yield / theoretical yield) × 100%
percent yield = (4.65g / 6.578g) × 100% ≈ 70.75%

Therefore, the percent yield of copper in this reaction is approximately 70.75%.