balance the equation
2CuO+C>>2Cu + CO2
each mole of carbon yields 2mmoles Cu
how many moles of Copper is 63.5 grams?
answer: 63.5/63.5=1mole
so you need half that number of moles of Carbon, or .5 moles of carbon, or six grams.
Ik this has been answered, but doesn't the copper oxide have something to do with it? And could you explain it in simpler terms? I'm just confused because you said forget the copper oxide, but your using it somehow, what puzzles me is the carbon dioxide, because the copper oxide plus the carbon make the copper, but also the carbon dioxide. Could you explain it simpler and explain how you got that answer (simpler)
Copper Oxide + Carbon -> Copper + Carbon Dioxide
Calculate the mass of carbon needed to obtain 63.5g of copper from 79.5g of Copper Oxide
2 answers
2CuO + C ==> 2Cu + CO2
1. I ignored the CuO because the problem states you want 63.5 g Cu. How many mols Cu is that? mols Cu = grams/atomic mass = 63.5/63.5 = 1 mol Cu is what you want.
2. If it comes from the equation above then 2 mols Cu is obtained for every mol C or 1/2 mol C for 1 mol Cu. So 1 mol Cu I will get from 1/2 mol C.
3. Since g = mols x atomic mass, then g C = mols C x atomic mass C = 0.5 x 12 = 6 grams C.
4. Note I didn't use CuO anywhere except in the equation. If it follows all of the laws of chemistry, knowing Cu I want will tell me how much CuO I need, how much C I need, and how much CO2 will be produced. But the problem didn't ask anything about CO2, or CuO so I didn't use them anywhere.
5. There COULD be a complicating factor IF (and only IF) the CuO would not give the amount of Cu you want in the problem BUT 1 mol CuO is 63.5 + 16 = 79.5; however, in that case I would have pointed out that it couldn't be done; i.e., 60 g CuO for example could not produce 63.5 g Cu so we were spinning our wheels.
To work a simple stoichiometry problem (as opposed to a limiting reagent problem) you need four steps.
1. Write and balance the equation.
2. Convert grams of what you have to mols. mols = grams/atomic mass in this case or grams/molar mass in the case of molecules. That's the 63.5/63.5 = 1 mol Cu step above.
3. Using the coefficients in the balanced equation, convert mols of what you have (in this case mols Cu) to mols of what you want (in this case C). That's the 1 mol Cu x (1 mol C/2 mols Cu) = 1 x 1/2 = 0.5 mols C
4. Then convert mols of what you want (the C) to grams. g = mols x atomic mass = 0.5 x 12 = 6.
These four steps will take care of any non-limiting reagent problem you have.
1. I ignored the CuO because the problem states you want 63.5 g Cu. How many mols Cu is that? mols Cu = grams/atomic mass = 63.5/63.5 = 1 mol Cu is what you want.
2. If it comes from the equation above then 2 mols Cu is obtained for every mol C or 1/2 mol C for 1 mol Cu. So 1 mol Cu I will get from 1/2 mol C.
3. Since g = mols x atomic mass, then g C = mols C x atomic mass C = 0.5 x 12 = 6 grams C.
4. Note I didn't use CuO anywhere except in the equation. If it follows all of the laws of chemistry, knowing Cu I want will tell me how much CuO I need, how much C I need, and how much CO2 will be produced. But the problem didn't ask anything about CO2, or CuO so I didn't use them anywhere.
5. There COULD be a complicating factor IF (and only IF) the CuO would not give the amount of Cu you want in the problem BUT 1 mol CuO is 63.5 + 16 = 79.5; however, in that case I would have pointed out that it couldn't be done; i.e., 60 g CuO for example could not produce 63.5 g Cu so we were spinning our wheels.
To work a simple stoichiometry problem (as opposed to a limiting reagent problem) you need four steps.
1. Write and balance the equation.
2. Convert grams of what you have to mols. mols = grams/atomic mass in this case or grams/molar mass in the case of molecules. That's the 63.5/63.5 = 1 mol Cu step above.
3. Using the coefficients in the balanced equation, convert mols of what you have (in this case mols Cu) to mols of what you want (in this case C). That's the 1 mol Cu x (1 mol C/2 mols Cu) = 1 x 1/2 = 0.5 mols C
4. Then convert mols of what you want (the C) to grams. g = mols x atomic mass = 0.5 x 12 = 6.
These four steps will take care of any non-limiting reagent problem you have.