250.0 ml of 0.250 m calcium chloride is mixed with 440.0 ml of 0.155 m sodium hydroxide and a precipitation reaction occurs. what is present at the end of the reaction? (report the amount of solid(s) formed in grams, and the concentration of products and left-over reagents in m)

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