what volume of 0.250 M calcium chloride needs to be added to sodium phosphate to produce 35.0 g of calcium phosphate ?
This is a stoichiometry problem. Here is a worked example of same; just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the volume of 0.250 M calcium chloride needed, we first need to calculate the moles of calcium phosphate required.
1. Find the molar mass of calcium phosphate (Ca3(PO4)2).
The molar mass of calcium (Ca) is 40.08 g/mol.
The molar mass of phosphorus (P) is 30.97 g/mol.
The molar mass of oxygen (O) is 16.00 g/mol.
So, the molar mass of calcium phosphate (Ca3(PO4)2) is:
(3 * 40.08) + (2 * (30.97 + 4 * 16.00))
= 120.24 + (2 * (30.97 + 64.00))
= 120.24 + (2 * 94.97)
= 120.24 + 189.94
= 310.18 g/mol.
2. Convert the mass of calcium phosphate to moles.
Mass = 35.0 g.
Moles = Mass / Molar mass
Moles = 35.0 g / 310.18 g/mol
Moles ≈ 0.113 mol.
3. Use the balanced chemical equation to find the moles of calcium chloride needed.
The balanced chemical equation for the reaction between calcium chloride (CaCl2) and sodium phosphate (Na3PO4) to form calcium phosphate (Ca3(PO4)2) is:
3CaCl2 + 2Na3PO4 → Ca3(PO4)2 + 6NaCl.
From the equation, we can see that one mole of calcium phosphate is formed from three moles of calcium chloride.
Moles of calcium chloride needed = 0.113 mol * (3 mol CaCl2 / 1 mol Ca3(PO4)2)
Moles of calcium chloride needed ≈ 0.339 mol.
4. Calculate the volume of 0.250 M calcium chloride needed.
Molarity (M) = Moles / Volume (L)
Volume (L) = Moles / Molarity
Volume (L) = 0.339 mol / 0.250 mol/L
Volume (L) ≈ 1.356 L.
Therefore, approximately 1.356 liters (or 1356 mL) of 0.250 M calcium chloride needs to be added to sodium phosphate to produce 35.0 g of calcium phosphate.
To find the volume of 0.250 M calcium chloride needed to produce 35.0 g of calcium phosphate, we need to first calculate the number of moles of calcium phosphate and then use stoichiometry to determine the amount of calcium chloride required.
Step 1: Calculate the moles of calcium phosphate
To determine the number of moles of calcium phosphate, we use the given mass and the molar mass of calcium phosphate, which is 310.18 g/mol.
Moles of calcium phosphate = mass of calcium phosphate / molar mass of calcium phosphate
Moles of calcium phosphate = 35.0 g / 310.18 g/mol
Step 2: Determine the stoichiometry between calcium phosphate and calcium chloride
The balanced equation for the reaction between calcium phosphate and calcium chloride is:
3 CaCl2 + 2 Na3PO4 → Ca3(PO4)2 + 6 NaCl
From the balanced equation, we can see that 3 moles of calcium chloride are required to produce 1 mole of calcium phosphate.
Step 3: Calculate the volume of calcium chloride
To find the volume of calcium chloride, we need to use the molarity of the calcium chloride solution and the number of moles of calcium chloride required.
Volume (L) of calcium chloride = moles of calcium chloride / molarity of calcium chloride
Since we already know the moles of calcium phosphate, we can calculate the moles of calcium chloride required by multiplying the moles of calcium phosphate by the stoichiometric ratio from the balanced equation (3 moles of CaCl2 per 1 mole of Ca3(PO4)2):
Moles of calcium chloride = 3 * moles of calcium phosphate
Now we can substitute this value in the formula to calculate the volume:
Volume (L) of calcium chloride = 3 * moles of calcium phosphate / molarity of calcium chloride
Note: Make sure to convert the molarity of calcium chloride to moles per liter (mol/L) to match the units of the moles of calcium chloride.
Finally, you can substitute the values into the equation to find the volume of calcium chloride solution that needs to be added.