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

hey there!

boiling point is the temperature at which the vapor pressure of the liquid equals atmospheric pressure. the vapor pressure of solvent molecules is decreased when a solute is added, so a higher temperature is required to increase the number of solvent molecules in the gas phase above the liquid.

at the freezing point, the vapor pressures of the solid and liquid are equal; a lower temperature is needed to reduce the number of solvent particles above the liquid.

hope this you

tobey