Copper sulfide is formed when copper and sulfur are heated together. In this reaction, 132 g of copper reacts with 53 g of sulfur. After the reaction is complete, 9 g of sulfur remains unreacted. What is the mass of copper sulfide formed?
To determine the mass of copper sulfide formed, we need to calculate the moles of copper and sulfur reacted and compare them to their stoichiometric ratio.
1. Calculate the moles of copper and sulfur reacted.
- Moles of copper (Cu):
- Mass of copper = 132 g
- Molar mass of copper (Cu) = 63.55 g/mol (from the periodic table)
- Moles of copper (Cu) = mass of copper / molar mass of copper
= 132 g / 63.55 g/mol
= 2.077 mol (rounded to three decimal places)
- Moles of sulfur (S):
- Mass of sulfur = (53 g - 9 g) = 44 g
- Molar mass of sulfur (S) = 32.06 g/mol (from the periodic table)
- Moles of sulfur (S) = mass of sulfur / molar mass of sulfur
= 44 g / 32.06 g/mol
= 1.373 mol (rounded to three decimal places)
2. Determine the stoichiometric ratio between copper and sulfur in the reaction.
The balanced chemical equation for the reaction is:
Cu + S -> CuS
From the equation, we can see that 1 mole of copper reacts with 1 mole of sulfur to form 1 mole of copper sulfide.
3. Calculate the moles of copper sulfide formed.
Since the stoichiometric ratio between copper and sulfur is 1:1, the moles of copper sulfide formed will be equal to the moles of copper or sulfur (whichever is smaller). In this case, the moles of sulfur (S) are smaller.
Moles of copper sulfide = Moles of sulfur (S) = 1.373 mol
4. Calculate the mass of copper sulfide formed.
- Molar mass of copper sulfide (CuS):
- Molar mass of copper (Cu) = 63.55 g/mol
- Molar mass of sulfur (S) = 32.06 g/mol
- Molar mass of copper sulfide (CuS) = Molar mass of copper (Cu) + Molar mass of sulfur (S)
= 63.55 g/mol + 32.06 g/mol
= 95.61 g/mol
- Mass of copper sulfide formed = Moles of copper sulfide * Molar mass of copper sulfide (CuS)
= 1.373 mol * 95.61 g/mol
= 131.35 g (rounded to two decimal places)
Therefore, the mass of copper sulfide formed is approximately 131.35 g.
To find the mass of copper sulfide formed, we need to determine the amount of sulfur that reacted and the amount of copper that reacted. We will then compare these amounts to find the limiting reactant.
First, let's calculate the amount of sulfur that reacted:
Given:
Mass of sulfur used = 53 g
Mass of unreacted sulfur = 9 g
Amount of sulfur reacted = Mass of sulfur used - Mass of unreacted sulfur
Amount of sulfur reacted = 53 g - 9 g = 44 g
Next, let's calculate the amount of copper that reacted using the molar ratio between copper and sulfur. The balanced chemical equation for this reaction is:
Cu + S -> CuS
From the equation, we can see that 1 mole of sulfur reacts with 1 mole of copper to form 1 mole of copper sulfide.
Molar mass of copper = 63.5 g/mol
Molar mass of sulfur = 32.1 g/mol
Using the molar masses, we can convert the mass of sulfur reacted into moles of sulfur:
Moles of sulfur reacted = Mass of sulfur reacted / Molar mass of sulfur
Moles of sulfur reacted = 44 g / 32.1 g/mol ≈ 1.37 mol
Since the molar ratio between copper and sulfur is 1:1, the moles of sulfur reacted will be the same as the moles of copper reacted.
Now, let's calculate the mass of copper that reacted:
Moles of copper reacted = Moles of sulfur reacted = 1.37 mol
Mass of copper reacted = Moles of copper reacted x Molar mass of copper
Mass of copper reacted = 1.37 mol x 63.5 g/mol ≈ 87 g
Finally, to find the mass of copper sulfide formed, we need to subtract the mass of sulfur remaining from the total mass of copper and sulfur initially used:
Mass of copper sulfide formed = Total mass of copper and sulfur used - Mass of remaining sulfur
Mass of copper sulfide formed = 132 g + 53 g - 9 g = 176 g
Therefore, the mass of copper sulfide formed is 176 g.
Copper (Cu) reacts with sulfur (S) to form copper sulfide as shown in the equation. A scientist adds 12.7 grams of Cu to 3.2 grams of S to start the reaction. In this reaction, all of the copper and all of the sulfur react. The scientist needs to determine how many grams of copper sulfide the reaction will produce. In 1–2 sentences, explain how to calculate the amount of copper sulfate this reaction will produce. Use the law of conservation of mass in the explanation.
This is a limiting reagent (LR) problem but they tell you that you have some S left over which means the LR is Cu.
2Cu + S ==> Cu2S
Follow the HgO problem to find mols Cu used up (all of it) and mols Cu2S formed (Note: the 9 g S unused is not use OTHER than so that you know Cu is the limiting reagent.)