If 0.0142 M aqueous Cl- is produced stoichiometrically according to the balanced equation in a reaction solution with a total volume of 1480 mL, how many mol of solid Ca3(PO4)2 are produced?
3 CaCl2(aq) + 2 Na3PO4(aq) → Ca3(PO4)2(s) + 6 NaCl(aq)
How many moles Cl^- are produced? That will be 0.0142M x 1.480L = ?
Convert that to moles Ca3(PO4)2.
?mol NaCl x (1 mol Ca3(PO4)2/6 moles NaCl) = ?mols NaCl x (1/6) = x mols Ca3(PO4)2.
To find out how many moles of solid Ca3(PO4)2 are produced, we can use the stoichiometry of the balanced equation. In the equation, it states that for every 3 moles of CaCl2, we get 1 mole of Ca3(PO4)2.
First, let's convert the given volume of the reaction solution from milliliters (mL) to liters (L). We divide the volume by 1000:
1480 mL = 1480/1000 L = 1.48 L
Now, we need to calculate the number of moles of Cl- ions produced. Since Cl- is a product in the reaction, and there are no coefficients in front of Cl- in the balanced equation, we know that the number of moles of Cl- ions produced is equal to the number of moles of Ca3(PO4)2 produced.
Next, we calculate the moles of Cl- ions using the given concentration (0.0142 M) and total volume (1.48 L):
Moles of Cl- ions = concentration × volume
Moles of Cl- ions = 0.0142 M × 1.48 L
Moles of Cl- ions = 0.0209 moles
Since the stoichiometry of the balanced equation tells us that for every 3 moles of CaCl2, we get 1 mole of Ca3(PO4)2, we can set up a proportion:
3 moles of CaCl2 = 1 mole of Ca3(PO4)2
0.0209 moles of CaCl2 = x moles of Ca3(PO4)2
Solving for x (moles of Ca3(PO4)2), we find:
x = (0.0209 moles × 1 mole of Ca3(PO4)2) / 3 moles of CaCl2
x = 0.00697 moles
Therefore, approximately 0.00697 moles of solid Ca3(PO4)2 are produced in the reaction.