science

Use the chemical equation to complete the activity.

2Cu+S→Cu2S
Copper (Cu) reacts with sulfur (S) to form copper sulfide as shown in the chemical equation. A scientist adds 4 grams of Cu to 2 grams of S to start the reaction. At the end of the experiment, she has 6 grams of product. In one to two sentences, explain what would happen if the scientist increased the amount of copper to 6 grams, and why.

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  1. I have news for you. It will not form 6 grams Cu2S. It will be more like 5 g Cu2S. It's like this and Cu is the limiting reagent.
    mols Cu initially = grams/atomic mass = 4/63.5 = 0.0630
    mols Cu2S from this would be 0.0630/2 = 0.0315 and grams Cu2S formed will be 0.0315 x 159 g Cu2S/mol = 5.00. OR if we go by the S route that is
    mols S = 2/32 = 0.0625 mols Cu2S from S will be 0.0625 x 159 g Cu2S/mol = 9.94 g Cu2S. It should be obvious that both answers (5.00 g Cu2S and 9.94 g Cu2S can't be right. In limiting reagent problems (LR) the SMALLER value is always the correct value. So Cu is the LR, you will form 5.00 g Cu2S and have an excess of S left over. I suspect that the author of the problem is applying the Law of Conservation of Mass, reasoning that 4g + 2 g = 6g but that's not the way this works in this problem BECAUSE this is a LR problem. It is true that you will get 5.00 g Cu2S, you will have 1.00 g S that didn't react, and the total will be 6 grams but that's not 6 g of product. That's 5 g product + 1 g of contamination. Now for the second part of the question.

    Now we have 6 g Cu and 2 g S.
    mols Cu = 6/63.5 = 0.0944
    mols S = 2/32 = 0.0625
    0.0944 mols Cu will form (with an excess of S) 0.0472 mols Cu2S
    0.0625 mols S will form (with an excess of Cu) 0.0625 mols Cu2S.
    So Cu still is the LR, 0.0472 mols Cu2S will form and that is 0.0472 x 159 = 7.50 g Cu2S. There will be 0.5 g S not reacted so the total is 6+2 = 8g. To be honest about it I don't know exactly what the author of the question is trying to prove. Hope this helps.

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    DrBob222
  2. @DrBob222 can you give me a shorter answer plz

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  3. I suspect that something went wrong in making up this problem. I was perplexed because it isn't obvious "what happened"? When I first looked at the problem I thought that in one case Cu would be the LR and in the other it would be S and that would make a discernible difference. But that isn't the case. Cu is the LR is both cases. So the only change I can see is that using more Cu produces more Cu2S. As to why that's because more Cu, in the presence of sufficient S, will produce more Cu2S.
    That section in bold is a shorter version. which is what you want; however, I think you had better understand exactly what is going on because I can see you getting a lot of flack over your answer. Hang in there though because what I've written is correct. I know what I'm doing. I would be interested in a follow up to let us know what kind of flack you get.

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    DrBob222
  4. dang i wanted to know if u got it right......heheeee

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  5. bruh thats not even regular science thats like rocket science or something

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  6. Wait who likes wayv!?

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