Sodium phosphate reacts with copper (II) chloride in a precipitation
reaction. What mass (in grams) of the precipitate will form when 325 mL of 0.5000 molar sodium phosphate reacts with 41.5 cL of 0.6000 molar copper (II) chloride.
This is a limiting reagent problem. How do I know? Because BOTH reactants are given.
1. Write the equation and balance it.
2. Convert Na3PO4 to moles. moles = M x L.
3. Convert CuCl2 to moles. same process.
4. Using the coefficients in the balanced equation, convert moles Na3PO4 to moles of the product.
5. Same process, convert moles CuCl2 to moles of the product.
6. It is quite likely that the answers from steps 4 and 5 will not agree which means one of them is wrong. The correct answer, in limiting reagent problems, is ALWAYS the smaller one and the reagent producing that number is the limiting reagent.
7. Now convert moles of the value from step 7 to grams. g = moles x molar mass.
To find the mass of the precipitate formed, we need to determine the limiting reactant between sodium phosphate and copper (II) chloride. The limiting reactant is the one that is completely consumed and determines the amount of product formed.
First, let's convert the given volumes into liters:
325 mL = 325/1000 = 0.325 L (sodium phosphate)
41.5 cL = 41.5/100 = 0.415 L (copper (II) chloride)
Now, let's calculate the number of moles of each reactant:
moles of sodium phosphate = volume (in liters) x molarity
moles of sodium phosphate = 0.325 L x 0.5000 mol/L = 0.1625 mol
moles of copper (II) chloride = volume (in liters) x molarity
moles of copper (II) chloride = 0.415 L x 0.6000 mol/L = 0.249 mol
Next, we need to determine the stoichiometry of the reaction. The balanced equation for the reaction between sodium phosphate (Na3PO4) and copper (II) chloride (CuCl2) is:
2 Na3PO4 + 3 CuCl2 → 6 NaCl + Cu3(PO4)2
From the balanced equation, we can see that 2 moles of sodium phosphate react with 3 moles of copper (II) chloride, forming 1 mole of copper (II) phosphate.
Now, let's calculate the number of moles of copper (II) phosphate formed:
moles of copper (II) phosphate = (moles of copper (II) chloride) x (1 mole of copper (II) phosphate / 3 moles of copper (II) chloride)
moles of copper (II) phosphate = 0.249 mol x (1/3) = 0.083 mol
Finally, we can calculate the mass of the copper (II) phosphate precipitate:
mass of copper (II) phosphate = moles of copper (II) phosphate x molar mass of copper (II) phosphate
The molar mass of copper (II) phosphate (Cu3(PO4)2) is:
(3 x atomic mass of copper) + (2 x (atomic mass of phosphorus + 4 x atomic mass of oxygen))
Calculating the molar mass:
(3 x 63.55 g/mol) + (2 x (31.00 g/mol + 4 x 16.00 g/mol)) = 380.13 g/mol
Finally, we can calculate the mass of the precipitate:
mass of copper (II) phosphate = 0.083 mol x 380.13 g/mol = 31.52 g
Therefore, approximately 31.52 grams of copper (II) phosphate will form as the precipitate.
To find the mass of the precipitate formed, we need to determine the limiting reactant first.
Step 1: Convert the given volumes to liters.
325 mL = 325/1000 = 0.325 L
41.5 cL = 41.5/100 = 0.415 L
Step 2: Calculate the number of moles of each reactant.
Moles of sodium phosphate = volume (L) × molarity
= 0.325 L × 0.5000 mol/L
= 0.1625 mol
Moles of copper (II) chloride = volume (L) × molarity
= 0.415 L × 0.6000 mol/L
= 0.249 mol
Step 3: Determine the mole ratio of the reactants.
From the balanced chemical equation, we can see that the mole ratio of sodium phosphate to copper (II) chloride is 2:3. Since the mole ratio is higher for copper (II) chloride, it will be the limiting reactant.
Step 4: Calculate the moles of copper (II) chloride reacting.
Moles of copper (II) chloride = (moles of sodium phosphate) × (3/2)
= 0.1625 mol × (3/2)
= 0.2438 mol
Step 5: Convert the moles of copper (II) chloride to mass.
To do this, we need the molar mass of copper (II) chloride, which is (63.5 g/mol + 2 × 35.5 g/mol) = 134.5 g/mol.
Mass of copper (II) chloride = moles × molar mass
= 0.2438 mol × 134.5 g/mol
= 32.8 g
Therefore, 32.8 grams of the precipitate will form when 325 mL of 0.5000 M sodium phosphate reacts with 41.5 cL of 0.6000 M copper (II) chloride.