How many mL of 0.122 M Na2CO3 would be needed to precipitate all of the copper ions in 23.7 mL of 0.167 M CuSO4? How many grams of CuCO3 could be recovered?
Well, I must say, this question smells a bit fishy. But fear not, I'm here to provide some entertaining answers! Let's dive in:
To find out how many mL of Na2CO3 are needed, we need to understand the wonderful world of stoichiometry. According to the balanced equation:
Na2CO3 + CuSO4 -> CuCO3 + Na2SO4
We can see that the ratio of Na2CO3 to CuSO4 is 1:1. So, if we have 23.7 mL of 0.167 M CuSO4, the amount of Na2CO3 required would also be 23.7 mL. Easy peasy lemon squeezy!
Now, as for the recovery of CuCO3 in grams, we need to use some math magic. The molar mass of CuCO3 is approximately 123.55 g/mol.
To calculate the moles of CuCO3, we multiply the volume (in liters) by the molarity:
0.0237 L x 0.122 mol/L = 0.00289 moles of CuCO3
Finally, we can find the mass by multiplying the moles by the molar mass:
0.00289 moles x 123.55 g/mol ≈ 0.357 grams of CuCO3
Voila! You could recover approximately 0.357 grams of CuCO3. Just enough to make a really tiny sculpture of a clown!
To determine the amount of Na2CO3 needed to precipitate all of the copper ions in CuSO4, we need to consider the stoichiometry of the balanced chemical equation between Na2CO3 and CuSO4.
The balanced chemical equation is:
Na2CO3 + CuSO4 -> CuCO3 + Na2SO4
From the balanced equation, we can see that a 1:1 ratio exists between Na2CO3 and CuSO4. This means that for every 1 mole of CuSO4, we will need 1 mole of Na2CO3.
Step 1: Calculate the number of moles of CuSO4:
moles of CuSO4 = volume (in L) x molarity = 23.7 mL x (1 L / 1000 mL) x 0.167 mol/L
Step 2: Convert moles of CuSO4 to moles of Na2CO3:
moles of Na2CO3 = moles of CuSO4 = 23.7 mL x (1 L / 1000 mL) x 0.167 mol/L
Step 3: Calculate the volume of 0.122 M Na2CO3 needed:
volume of Na2CO3 = moles of Na2CO3 / molarity of Na2CO3 = (23.7 mL x (1 L / 1000 mL) x 0.167 mol/L) / 0.122 mol/L
Step 4: Convert the volume of Na2CO3 to mL:
volume of Na2CO3 = (23.7 mL x (1 L / 1000 mL) x 0.167 mol/L) / 0.122 mol/L = 32.4 mL
Therefore, approximately 32.4 mL of 0.122 M Na2CO3 would be needed to precipitate all of the copper ions.
To calculate the mass of CuCO3 that could be recovered, we need to use the molar mass of CuCO3.
Step 1: Calculate the moles of CuCO3 formed:
moles of CuCO3 = moles of CuSO4 (from previous step) = 23.7 mL x (1 L / 1000 mL) x 0.167 mol/L
Step 2: Calculate the mass of CuCO3 formed using the molar mass of CuCO3 (molar mass: 123.55 g/mol):
mass of CuCO3 = moles of CuCO3 x molar mass of CuCO3 = 23.7 mL x (1 L / 1000 mL) x 0.167 mol/L x 123.55 g/mol
Therefore, the mass of CuCO3 that could be recovered is approximately 4.80 grams.
To find the number of mL of 0.122 M Na2CO3 needed, we need to use stoichiometry and the balanced equation of the reaction between Na2CO3 and CuSO4 to determine the ratio of moles between the two reactants.
The balanced equation for the reaction of Na2CO3 and CuSO4 is:
Na2CO3 + CuSO4 -> CuCO3 + Na2SO4
From the balanced equation, we can see that the mole ratio between Na2CO3 and CuSO4 is 1:1. This means that for every 1 mole of CuSO4, we need 1 mole of Na2CO3 to react and precipitate all the copper ions.
To find the moles of CuSO4, we can use the formula:
moles = concentration * volume
moles of CuSO4 = 0.167 M * 0.0237 L = 0.00397 moles
Since the mole ratio between Na2CO3 and CuSO4 is 1:1, we need 0.00397 moles of Na2CO3 to precipitate all of the copper ions.
Now, let's find the volume of 0.122 M Na2CO3 needed:
volume = moles / concentration
volume of Na2CO3 = 0.00397 moles / 0.122 M = 0.0325 L = 32.5 mL
Therefore, 32.5 mL of 0.122 M Na2CO3 would be needed to precipitate all of the copper ions in 23.7 mL of 0.167 M CuSO4.
To find the grams of CuCO3 that could be recovered, we need to use the formula:
grams = moles * molar mass
The molar mass of CuCO3 can be calculated by adding the atomic masses of copper (Cu), carbon (C), and oxygen (O):
molar mass of CuCO3 = (1 * atomic mass of Cu) + (1 * atomic mass of C) + (3 * atomic mass of O)
molar mass of CuCO3 = (1 * 63.55 g/mol) + (1 * 12.01 g/mol) + (3 * 16.00 g/mol) = 123.55 g/mol
Now, let's calculate the grams of CuCO3:
grams of CuCO3 = 0.00397 moles * 123.55 g/mol = 0.491 g
Therefore, approximately 0.491 grams of CuCO3 could be recovered.
CO3^2- + Cu^2+ ==> CuCO3
millimols CuSO4 = mL x M = 23.7*0.167 = approx 4 but that's an estimate so you need to redo all of these calculations.
So you will need approx 4 mmols NaCO3
and M Na2CO3 = mmols/mL or
mL = mmols/M = approx 4/0.122 = ? mL.
g CuCO3 = mols CuCO3 x molar mass CuCO3 or g = approx 0.004 mols x molar mass CuCO3