When copper is heated with an excess of sulfur, copper(I) sulfide is formed. In a given experiment, 0.0970 moles of copper was heated with excess sulfur to yield 1.76 g copper(I) sulfide. What is the percent yield?

The percentage yield of the experiment obtained by the reaction of 0.0970 mole of copper with excess sulfurs is 22.8%

We'll begin by calculating the number of mole of Cu₂S produced from the reaction. This can be obtained as follow:

2Cu + S —> Cu₂S

From the balanced equation above,

2 moles of Cu reacted to produce 1 mole of Cu₂S.

Therefore,

0.0970 mole of Cu will react to produce = = 0.0485 mole of Cu₂S.

Next, we shall determine the theoretical yield by calculating the mass of 0.0485 mole of Cu₂S.

Molar mass of Cu₂S = (63.5×2) + 32 = 159 g/mol

Mole of Cu₂S = 0.0485 mole

Mass = mole × molar mass

Mass of Cu₂S = 0.0485 × 159

Thus, the theoretical yield of Cu₂S is 7.7115 g

Finally, we shall determine the percentage yield of Cu₂S.

Actual yield = 1.76 g

Theoretical yield = 7.7115 g

Therefore, the percentage yield of the experiment is 22.8%

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Percent yield = 22.8 %

Explanation:

Step 1: Data given

Numbers of moles  copper = 0.0970 moles

Mass of copper(I) sulfide = 1.76 grams

Step 2: The balanced equation

2Cu + S ⇒ Cu2S

Step 3:  Calculate moles of Cu2S

For 2 moles Cu we need 1 mol S to produce 1 mol Cu2S

For 0.0970 moles Cu we'll hace 0.0970 / 2 = 0.0485 moles

Step 4: Calculate mass of Cu2S

Mass Cu2s = moles Cu2S * molar mass Cu2S

Mass Cu2S = 0.0485 moles * 159.16 g/mol

Mass Cu2S = 7.72 grams

Step 5: Calculate percent yield

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

Percent yield = (1.76 grams / 7.72 grams)*100%

Percent yield = 22.8 %

output?

explanation:

i'm assuming that it's c.) red bank..