Answer:
Solid: 2.52 g
Concentrations: [CaCl₂] = 0.041 M, [NaCl] = 0.100 M
Explanation:
When calcium chloride (CaCl₂) reacts with sodium hydroxide (NaOH), a double replacement reaction occurs, forming NaCl and Ca(OH)₂. NaCl is a soluble salt, and Ca(OH)₂ is a little soluble base, thus, Ca(OH)₂ will be the precipiate.
The balanced reaction equation is:
CaCl₂(aq) + 2NaOH(aq) → 2NaCl(aq) + Ca(OH)₂(s)
The number of moles of the reactants mixed are their volume multiplied by their concentration:
nCaCl₂ = 0.250 L * 0.25 mol/L = 0.0625 mol
nNaOH = 0.440 L*0.155 mol/L = 0.0682 mol
One of the reactants is limiting, and the other is in excess. Let's suppose that CaCl₂ is limiting, then, by the stoichiometry:
1 mol of CaCl₂ ------------- 2 moles of NaOH
0.0625 mol ------------ x
By a simple direct three rule:
x = 0.125 mol of NaOH
Because there's less NaOH than the value found, NaOH must be the limiting reactant and CaCl₂ is in excess. Thus, by the stoichiometry:
1 mol of CaCl₂ ------------- 2 moles of NaOH
x ------------- 0.0682 mol
By a simple direct three rule:
2x = 0.0682
x = 0.0341 mol of CaCl₂ reacts
The number of moles of CaCl₂ that remains is: 0.0625 - 0.0341 = 0.0284 mol. The final volume is 250.0 mL + 440.0 mL = 690. mL = 0.69 L
[CaCl₂] = 0.0284/0.69 = 0.041 M
For the solube product:
2 moles of NaOH ------------ 2 moles of NaCl
0.0682 mol ------------ x
x = 0.0682 mol of NaCl formed
[NaCl] = 0.0682/0.69 = 0.100 M
For the precipitate:
2 moles of NaOH ----------- 1 mol of Ca(OH)₂
0.0682 mol ---------- x
x = 0.0341 mol of Ca(OH)₂ formed
The molar of Ca(OH)₂ is 74.0 g/mol, and the mass is the number of moles multiplied by the molar mass:
mCa(OH)₂ = 0.0341*74 = 2.52 g
