What is the maximum mass of cobalt (II) phosphate that will precipitate from a solution prepared by adding an excess of K3PO4 solution to 100.0 mL of 5.0 M CoCI2?
Sorry for all the confusion with different names we are working in a group and its easier to keep track who's work is who's
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No problem at all! Let's work through the problem together step by step.
To find the maximum mass of cobalt (II) phosphate that will precipitate from the solution, we need to consider the stoichiometry of the reaction and the limiting reactant.
The balanced equation for the reaction between cobalt (II) chloride (CoCl2) and potassium phosphate (K3PO4) is:
3 CoCl2 + 2 K3PO4 -> Co3(PO4)2 + 6 KCl
From the balanced equation, we can see that 3 moles of cobalt (II) chloride react with 2 moles of potassium phosphate to form 1 mole of cobalt (II) phosphate.
First, we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed, thus determining the maximum amount of product that can be formed.
Given that we have an excess of K3PO4 solution, we need to calculate the number of moles of cobalt (II) chloride in the 100.0 mL of 5.0 M CoCl2 solution.
To do this, we can use the following equation:
moles = concentration (M) x volume (L)
First, convert the volume given to liters:
100.0 mL = 100.0 mL * (1 L / 1000 mL) = 0.100 L
Next, use the concentration and volume to determine the number of moles of cobalt (II) chloride:
moles of CoCl2 = 5.0 M * 0.100 L = 0.500 moles
Now, we can use stoichiometry to calculate the maximum amount of cobalt (II) phosphate that will precipitate.
From the balanced equation, we know that 3 moles of CoCl2 react with 1 mole of Co3(PO4)2. Therefore, the moles of Co3(PO4)2 formed will be:
moles of Co3(PO4)2 = (0.500 moles CoCl2) / 3 = 0.167 moles
Finally, we need to convert the moles of Co3(PO4)2 to grams using the molar mass of cobalt (II) phosphate.
The molar mass of cobalt (II) phosphate (Co3(PO4)2) is:
Co: 58.93 g/mol
P: 30.97 g/mol
O: 16.00 g/mol
molar mass of Co3(PO4)2 = (3 * 58.93 g/mol) + (2 * 30.97 g/mol) + (8 * 16.00 g/mol) = 457.82 g/mol
Finally, we can calculate the maximum mass of cobalt (II) phosphate:
mass = moles * molar mass = 0.167 moles * 457.82 g/mol = 76.39 grams
Therefore, the maximum mass of cobalt (II) phosphate that will precipitate from the solution is approximately 76.39 grams.