When a reaction mixture with a total volume of 1180 mL that is 0.0110 M aqueous NaCl was stoichiometrically produced as per the balanced equation, what volume (mL) of 0.0116 M aqueous Ca2+ was required?
3 CaCl2(aq) + 2 Na3PO4(aq) → Ca3(PO4)2(s) + 6 NaCl(aq)
Molar Mass (g)
NaCl 58.433
Ca2+ 40.078
Molar Volume (L)
22.400 at STP
Gas Constant
0.0821
To determine the volume of 0.0116 M aqueous Ca2+ required in the reaction, we need to use the stoichiometry of the balanced equation and the given concentration and volume of the NaCl solution.
1. Start by writing down the balanced equation and determining the stoichiometric ratio between NaCl and Ca2+. From the given equation:
3 CaCl2(aq) + 2 Na3PO4(aq) → Ca3(PO4)2(s) + 6 NaCl(aq)
The stoichiometric ratio between CaCl2 and NaCl is 3:6, which simplifies to 1:2. This means that for every 1 mole of CaCl2, we need 2 moles of NaCl.
2. Calculate the number of moles of NaCl in the given reaction mixture. We can use the concentration and volume of NaCl to find the moles:
Moles of NaCl = concentration * volume
= 0.0110 M * 1180 mL
= 0.0110 mol/L * 1.180 L
= 0.01298 moles
3. Determine the moles of Ca2+ required using the stoichiometry of the balanced equation. Since we know that the stoichiometric ratio between CaCl2 and NaCl is 1:2, the moles of Ca2+ required will be half of the moles of NaCl:
Moles of Ca2+ required = Moles of NaCl / 2
= 0.01298 moles / 2
= 0.00649 moles
4. Using the concentration and moles of Ca2+, we can now calculate the volume of the Ca2+ solution required:
Volume of Ca2+ solution = Moles of Ca2+ required / concentration
= 0.00649 moles / 0.0116 M
≈ 0.5577 L
= 557.7 mL (approximately)
Therefore, approximately 557.7 mL of the 0.0116 M aqueous Ca2+ solution is required.