Copper(II) sulfide reacts with oxygen to produce copper (II) oxide plus sulfur dioxide.

Suppose you start this reaction with 50.0 grams of copper(II) sulfide and 50.0 grams of oxygen and you actually produce 33.5 grams of copper(II) oxide.

Calculate the percent yield for the reaction.

COLORLESS.GASN2O4EXISTEQUILBRIUM

phosphorus (P4) is commercially prepared by heating a mixture of calcium phosphate (Ca3(PO4)2), sand (SiO2) and coke (C) in an elecric furnace. The process involves two reactions.

To calculate the percent yield of a reaction, you need to compare the actual yield (the amount of product obtained in a reaction) to the theoretical yield (the amount of product that would be obtained if the reaction went to completion according to stoichiometry).

First, let's calculate the theoretical yield of copper(II) oxide. We can do this by determining the limiting reactant, which is the reactant that is completely consumed in the reaction, thereby limiting the amount of product formed.

To find the limiting reactant, we need to calculate the number of moles of each reactant. We can use their molar masses and the given masses of reactants.

The molar mass of copper(II) sulfide (CuS) is the sum of the molar masses of copper (Cu) and sulfur (S):
CuS = atomic mass of Cu + atomic mass of S
= 63.55 g/mol + 32.07 g/mol
= 95.62 g/mol

The molar mass of oxygen (O2) is 2 times the atomic mass of oxygen (O):
O2 = 2 * atomic mass of O
= 2 * 16.00 g/mol
= 32.00 g/mol

Now we can calculate the number of moles for each reactant:

Number of moles of copper(II) sulfide = mass of copper(II) sulfide / molar mass of copper(II) sulfide = 50.0 g / 95.62 g/mol = 0.523 mol

Number of moles of oxygen = mass of oxygen / molar mass of oxygen = 50.0 g / 32.00 g/mol = 1.563 mol

Based on the balanced chemical equation, the reaction requires 1 mole of copper(II) sulfide and 1 mole of oxygen to produce 1 mole of copper(II) oxide. Therefore, the mole ratio between copper(II) sulfide and copper(II) oxide is 1:1.

Since the number of moles of copper(II) sulfide is less than the number of moles of oxygen, copper(II) sulfide is the limiting reactant. This means that all of the copper(II) sulfide will be consumed in the reaction, and the amount of copper(II) oxide formed will be determined by the stoichiometry of the reaction.

Using the mole ratio, we can now calculate the theoretical yield of copper(II) oxide:

Theoretical yield of copper(II) oxide = number of moles of copper(II) sulfide * molar mass of copper(II) oxide
= 0.523 mol * (63.55 g/mol + 16.00 g/mol)
= 0.523 mol * 79.55 g/mol
= 41.64 g

Now that we have the theoretical yield (41.64 g) and the actual yield (33.5 g), we can calculate the percent yield using the formula:

Percent yield = (actual yield / theoretical yield) * 100

Percent yield = (33.5 g / 41.64 g) * 100
= 0.8057 * 100
= 80.57%

Therefore, the percent yield for the reaction is approximately 80.57%.