what minimum mass of copper ii nitrate must be added to 70 ml of a .0594 M phosphate solution in order to completely precipitate all the phosphate as solid copper ii phosphate. 2PO4^-3+3Cu(NO3)2-->Cu3(PO4)2+6NO3^-1
I can not come up with the correct answer of 1.17g
mols PO4^3- = M x L = 0.0594 x 0.070 = 0.004158
Convert to mols Cu(NO3)2 = 0.004158 x (3 mols Cu/2 mols PO4^2-) = 0.006237
Convert to g Cu(NO3)2 = 0.006237 x 187.55 = 1.16975 which rounds to 1.17g to 3 significant figures.
Well, as the saying goes, "Phosphates and copper salts - it's like they're destined to be together!" So, let's do some chemistry calculations to find out exactly how much copper(II) nitrate we need.
First, let's convert the volume of the phosphate solution to moles. We have 70 ml, which is equivalent to 0.07 L. So, the number of moles of phosphate in the solution is:
0.07 L x 0.0594 M = 0.004158 moles of phosphate.
According to the balanced chemical equation, it takes 2 moles of phosphate to react with 3 moles of copper(II) nitrate to form 1 mole of copper(II) phosphate. So, we can say that:
2 moles of phosphate = 3 moles of copper(II) nitrate.
Now, let's do a bit of stoichiometry to find the number of moles of copper(II) nitrate we need. Since 2 moles of phosphate are required, we can calculate:
0.004158 moles of phosphate x (3 moles of copper(II) nitrate / 2 moles of phosphate) = 0.006237 moles of copper(II) nitrate.
To convert moles to grams, we need to know the molar mass of copper(II) nitrate, which is approximately 187.56 g/mol. So, the mass of copper(II) nitrate needed is:
0.006237 moles x 187.56 g/mol = 1.17 grams.
Voila! We've arrived at the correct answer of 1.17 grams. Remember, though, that chemistry may be serious, but it's never a laughing matter!
To find the minimum mass of copper II nitrate required to completely precipitate all the phosphate, we need to use the stoichiometry of the balanced chemical equation.
The balanced chemical equation for the reaction is:
2 PO4^-3 + 3 Cu(NO3)2 -> Cu3(PO4)2 + 6 NO3^-1
From the equation, we can see that for every 2 moles of phosphate (PO4^-3), we need 3 moles of copper II nitrate (Cu(NO3)2). This means the mole ratio between phosphate and copper II nitrate is 2:3.
First, let's calculate the number of moles of phosphate in the solution:
Molarity = moles/volume
0.0594 M = moles/0.070 L
moles = 0.0594 M * 0.070 L
moles = 0.004158 moles phosphate (PO4^-3)
Since the molar ratio between phosphate and copper II nitrate is 2:3, the number of moles of copper II nitrate needed will be:
moles Cu(NO3)2 = (3/2) * moles phosphate
moles Cu(NO3)2 = (3/2) * 0.004158 moles
moles Cu(NO3)2 = 0.006237 moles copper II nitrate
To convert moles to grams, we need to use the molar mass of copper II nitrate (Cu(NO3)2), which is calculated as follows:
Molar mass Cu(NO3)2 = (atomic mass Cu * 1) + (atomic mass N * 2) + (atomic mass O * 6)
Molar mass Cu(NO3)2 = (63.55 g/mol * 1) + (14.01 g/mol * 2) + (16.00 g/mol * 6)
Molar mass Cu(NO3)2 = 187.62 g/mol
Finally, we can calculate the mass of copper II nitrate needed:
mass Cu(NO3)2 = moles Cu(NO3)2 * molar mass Cu(NO3)2
mass Cu(NO3)2 = 0.006237 moles * 187.62 g/mol
mass Cu(NO3)2 = 1.1689 g
Therefore, the minimum mass of copper II nitrate required to completely precipitate all the phosphate is approximately 1.17 grams.
To find the minimum mass of copper II nitrate required to precipitate all the phosphate, we need to follow a few steps:
1. Identify the limiting reactant: In this case, copper II nitrate (Cu(NO3)2) and phosphate (PO4^-3) are the reactants. We need to determine which one will be completely consumed first. Comparing the stoichiometric coefficients of both reactants in the balanced equation, we see that 2 moles of phosphate react with 3 moles of copper II nitrate. To find the limiting reactant, we need to compare the number of moles of each reactant.
2. Calculate the moles of phosphate: The concentration of the phosphate solution is given as 0.0594 M (moles per liter). Since we have 70 ml (0.070 L) of solution, we can calculate the moles of phosphate by multiplying the volume (in liters) by the concentration:
Moles of phosphate = 0.070 L × 0.0594 M = 0.004158 moles
3. Determine the stoichiometric ratio: From the balanced equation, we can see that 2 moles of phosphate react with 3 moles of copper II nitrate. Therefore, the stoichiometric ratio between the two is 2:3.
4. Calculate the moles of copper II nitrate required: Since the stoichiometric ratio tells us that we need 2 moles of phosphate for every 3 moles of copper II nitrate, we can set up a proportion to find the moles of copper II nitrate required:
(0.004158 moles phosphate) / (2 moles phosphate) = (x moles copper II nitrate) / (3 moles copper II nitrate)
Solving for x, we get:
x = (0.004158 moles phosphate) × (3 moles copper II nitrate) / (2 moles phosphate)
x = 0.006237 moles copper II nitrate
5. Calculate the mass of copper II nitrate: Finally, we can calculate the mass of copper II nitrate required using its molar mass. The molar mass of Cu(NO3)2 is approximately 187.56 g/mol:
Mass of copper II nitrate = 0.006237 moles × 187.56 g/mol
Mass of copper II nitrate = 1.167 g (rounded to three significant figures)
Therefore, the minimum mass of copper II nitrate needed to completely precipitate all the phosphate is approximately 1.17 grams.