Calculate the mass of CuFeS2 needed in g to produce 600.0 g Cu if the yield for the conversion process is 82%.
Cu = 63.5 g/mol
Fe = 56 g/mol
2 S = 32 *2 = 64 g/ mol
so
CuFeS2 = 63.5 + 56 + 64 grams/ mol of CuFeS2
so if 100 % yield
(63.5+56+64) grams CuFeS2 for 63.5 grams Cu
so
(63.5+56+64 ) / ( 0.82* 63.5)
Damon, I think your calculation is to produce 1 g Cu. Multiply by 600 to produce 600 grams as per the problem.
Well, considering the yield of the conversion process is 82%, I guess we can say that 18% of the mass will go "poof" and disappear. So, let's calculate the amount of CuFeS2 we need by using a little math.
First, we need to find out what 18% of 600.0 g is.
18% of 600.0 g = (18/100) * 600.0 g = 108.0 g
So, 108.0 g of Cu will disappear in the conversion process, leaving us with 600.0 g - 108.0 g = 492.0 g of copper.
Now, the atomic mass of Cu is 63.546 g/mol, and the molar ratio between Cu and CuFeS2 is 1:1. This means that we'll need the same amount of CuFeS2 in grams as the molar mass of Cu, which is 63.546 g/mol.
Therefore, to produce 600.0 g of Cu, we'll need 63.546 g of CuFeS2.
I hope my calculations didn't leave you feeling "cu-rious"!
To calculate the mass of CuFeS2 needed, we can use the concept of stoichiometry. The balanced chemical equation for the conversion of CuFeS2 to Cu is:
CuFeS2 + O2 → Cu + FeO + SO2
From the balanced equation, we see that the molar ratio between CuFeS2 and Cu is 1:1. This means that 1 mole of CuFeS2 is required to produce 1 mole of Cu.
We need to convert the given mass of Cu to moles using its molar mass, which is 63.55 g/mol. The calculation will be:
Moles of Cu = (mass of Cu) / (molar mass of Cu)
= 600.0 g / 63.55 g/mol
= 9.44 mol Cu
Since the yield of the conversion process is 82%, we need to consider this in our calculation. The actual amount of CuFeS2 needed will be:
Moles of CuFeS2 = (moles of Cu) / (yield)
= 9.44 mol / 0.82
= 11.51 mol CuFeS2
Finally, to convert moles of CuFeS2 to grams, we multiply by its molar mass, which is 183.52 g/mol. The mass of CuFeS2 needed is:
Mass of CuFeS2 = (moles of CuFeS2) x (molar mass of CuFeS2)
= 11.51 mol x 183.52 g/mol
= 2115.39 g
Therefore, the mass of CuFeS2 needed to produce 600.0 g of Cu with a yield of 82% is approximately 2115.39 g.
To calculate the mass of CuFeS2 needed to produce 600.0 g of Cu, we need to consider the stoichiometry of the reaction and the yield of the conversion process.
The balanced chemical equation for the reaction is:
CuFeS2 + O2 → Cu + FeO + SO2
From the equation, we can see that the mole ratio between CuFeS2 and Cu is 1:1. This means that 1 mole of CuFeS2 produces 1 mole of Cu.
First, let's calculate the moles of Cu using the given mass of Cu:
moles of Cu = mass of Cu / molar mass of Cu
The molar mass of Cu (copper) is approximately 63.55 g/mol.
moles of Cu = 600.0 g / 63.55 g/mol = 9.44 mol
Since the yield of the conversion process is 82%, only 82% of the Cu will be produced. Therefore, we need to adjust the number of moles accordingly:
moles of Cu produced = moles of Cu * yield
moles of Cu produced = 9.44 mol * 0.82 = 7.73 mol
Since the stoichiometric ratio is 1:1 between CuFeS2 and Cu, the moles of CuFeS2 needed are also 7.73 mol.
Finally, to calculate the mass of CuFeS2 needed, we use the molar mass of CuFeS2 (copper iron sulfide). The molar mass of CuFeS2 is approximately 183.5 g/mol.
mass of CuFeS2 needed = moles of CuFeS2 * molar mass of CuFeS2
mass of CuFeS2 needed = 7.73 mol * 183.5 g/mol = 1417.88 g
Therefore, you would need approximately 1417.88 grams of CuFeS2 to produce 600.0 grams of Cu with an 82% yield in the conversion process.