Copper reacts with sulfur to form Copper(I)sulfide. 2Cu(s) + S(s) ----> Cu2S

What is the maximum number if grams Cu2S that can be formed when 80 grams of Copper reacts with 25.0 grams of sulfur.

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Total moles on this side = Total moles on that side. Whack them moles.
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Uhhh explain further, I dont think that works.
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Copper(I)sulfide. 2Cu(s) + S(s) ----> Cu2S
25.0 grams
80 grams + Copper(I)sulfide. 2Cu(s) + S(s) ----> Cu2S - 25.0 grams = jellyfish
its actually copper sulphate...

You have to divide the mass by the Mr to get the number of moles in each. Then work out the maximum number of moles formed by finding the Mr of Cu2S, then multiply the number of moles by the Mr to find the mass in grams.

Thats the loose method. I'm pretty certain its right but its the holidays and my brain isn't in gear.
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I didn't actually..
Last edited by HammettHead at Apr 3, 2008,
its actually copper sulphate...

You have to divide the mass by the Mr to get the number of moles in each. Then work out the maximum number of moles formed by finding the Mr of Cu2S, then multiply the number of moles by the Mr to find the mass in grams.

Thats the loose method. I'm pretty certain its right but its the holidays and my brain isn't in gear.

Copper Sulphate = CuSO4

You need oxygen to make it sulphate.
Oh god it's 5:30 in the morning and I'm actually bored enough to do this...

First you find the number of mols of each element, using n=M/m.

mol(Cu) = 63.54/80 = 0.79425mol
mol(S) = 32.07/25 = 1.2828mol

Now you find the limiting reagent of the reaction, by subbing in one of the values for mols and seeing if the other is in excess or not, so from the reaction:
2(mol Cu) = mol S
2(0.79425) = 1.5885mol
But we only have 1.2828mol of S, therefore it is the limiting reagent.

So now we sub in the number of mols of S and use the mol ratios to find the number of mols of Cu2S:
mol S = mol Cu2S = 1.2828mol

Now use the n=M/m formula again to find mass of Cu2S (note Molar mass of Cu2S is 159.15g/mol):
m = 159.15/1.2828=124.06 grams
yeah, barring a calculation error, e-ron has it right
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