13. Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as

follows:
Cu2S(s) + O2(g) „_ 2Cu(s) + SO2(g)
If the reaction of 0.540 kg of copper(I) sulfide with excess oxygen produces 0.140 kg of
copper metal, what is the percent yield?
A) 32.5%
B) 25.9%
C) 64.9%
D) 130%
E) 39.9%

If that reaction were to go to completion, one mole of Cu2S(weighing 159 g) would form 2 moles of Cu (weighing 127g). 0.540 kg of Cu2S would thus lead to production of

0.540 * (127/159) = 0.431 g of Cu.

The 0.140 g actually produced is 32.5% of that amount. That is the percent yield.

Convert 0.540 kg Cu2S to moles. Moles = grams/molar mass.

Using the coefficients in the balanced equation, convert moles Cu2S to moles Cu metal.

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

% yield = (0.140 kg/theoretical yield)*100 = ??
Be sure numerator and denominator are in the same units.

To calculate the percent yield, we need to compare the actual yield (0.140 kg of copper) with the theoretical yield (the maximum amount of copper that could be produced based on the stoichiometry of the balanced chemical equation).

First, we need to calculate the theoretical yield of copper.

According to the balanced chemical equation, 1 mole of Cu2S reacts to produce 2 moles of Cu.

1 mole of Cu2S has a molar mass of (2 x 63.55 g/mol) + 32.07 g/mol = 159.17 g/mol.

The number of moles of Cu2S can be calculated by dividing its mass by its molar mass:

Number of moles of Cu2S = 0.540 kg / 159.17 g/mol = 3.393 moles

From the stoichiometry of the balanced equation, we know that 1 mole of Cu2S produces 2 moles of Cu. Therefore, the number of moles of Cu produced would be:

Number of moles of Cu = 3.393 moles x 2 = 6.786 moles

The molar mass of Cu is 63.55 g/mol. Multiplying the number of moles of Cu by its molar mass gives us the theoretical yield of Cu:

Theoretical yield of Cu = 6.786 moles x 63.55 g/mol = 431.16 g

Now, we can calculate the percent yield by dividing actual yield by theoretical yield and multiplying by 100:

Percent yield = (0.140 kg / 431.16 g) x 100 = 32.5%

Therefore, the percent yield is approximately 32.5%. Hence, the correct answer choice is A) 32.5%.

To find the percent yield, we need to compare the actual yield (0.140 kg) to the theoretical yield (the maximum amount of copper that can be produced based on the given amount of copper(I) sulfide).

First, we need to calculate the molar mass of copper(I) sulfide (Cu2S) and copper (Cu) using the periodic table:

- Molar mass of Cu2S: (2 x atomic mass of Cu) + atomic mass of S
= (2 x 63.55 g/mol) + 32.07 g/mol
= 159.17 g/mol

- Molar mass of Cu: 63.55 g/mol

Next, we need to calculate the number of moles of copper(I) sulfide and copper:

- Moles of Cu2S: mass of Cu2S / molar mass of Cu2S
= 0.540 kg / 159.17 g/mol
= 0.00339 mol

- Moles of Cu: mass of Cu / molar mass of Cu
= 0.140 kg / 63.55 g/mol
= 0.00220 mol

According to the balanced chemical equation, 1 mole of Cu2S reacts to form 2 moles of Cu. Therefore, the theoretical yield of Cu can be calculated:

- Theoretical yield of Cu: 2 x moles of Cu2S
= 2 x 0.00339 mol
= 0.00678 mol

Now, we can calculate the percent yield:

- Percent yield: (actual yield / theoretical yield) x 100%
= (0.00220 mol / 0.00678 mol) x 100%
= 32.46%

The closest answer choice to 32.46% is 32.5%, which is option A).