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

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.

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.

To calculate the mass of carbon needed, we start by understanding the chemical reaction provided:

Copper Oxide + Carbon -> Copper + Carbon Dioxide

In this reaction, Copper Oxide (CuO) reacts with Carbon (C) to produce Copper (Cu) and Carbon Dioxide (CO2).

To find the mass of carbon needed, we need to consider the law of conservation of mass, which states that mass cannot be created or destroyed in a chemical reaction. This means that the total mass of the reactants must be equal to the total mass of the products.

Given that we have 79.5g of Copper Oxide, we can start by determining the molar mass of Copper Oxide.

Copper (Cu) has an atomic mass of 63.5 g/mol, and Oxygen (O) has an atomic mass of 16.0 g/mol. Therefore, the molar mass of Copper Oxide (CuO) can be calculated as follows:

Molar Mass of CuO = (63.5 g/mol) + (16.0 g/mol)
= 79.5 g/mol

This means that 79.5g of Copper Oxide is equivalent to one mole of Copper Oxide, which weighs 79.5 grams.

To determine the mass of carbon needed to obtain 63.5g of copper, we can set up a ratio based on the balanced equation:

1 mole of CuO reacts with 1 mole of C to produce 1 mole of Cu.

Since the molar mass of Carbon (C) is 12.0 g/mol, we can calculate the moles of Copper using the given mass of Copper (63.5 g) as follows:

Moles of Cu = (63.5 g) / (63.5 g/mol)
= 1 mole

Since we have a 1:1 mole ratio between Copper (Cu) and Carbon (C), this means we need 1 mole of Carbon to produce 1 mole of Copper.

To find the mass of Carbon needed, we can convert the moles of Carbon to grams using its molar mass:

Mass of C = (1 mole) x (12.0 g/mol)
= 12.0 g

Therefore, the mass of carbon needed to obtain 63.5g of copper from 79.5g of Copper Oxide is 12.0 grams.