[mechanick]

I have an assignment to make road salt (CaCl₂) less effective.  I have had a lot of trouble finding any substance that would work.  My only idea so far is:
CaCl_2(aq) + PbC_2(aq)   CaC_2(s) + PbCl_2(s)
CaC_2(s) + 2H₂O_(l)   C₂H_2(g) + Ca(OH)_2(s)

Would this reaction work?

Also, can anyone help me to find a compound that would react with CaCl₂ to make it less effective?  Thanks for the help.

[408]

Lead carbide?  You are not going to find that commonly, nor will it be soluble in water.

To make it less effective look at how how CaCl2 suppresses the melting point of ice.  Then look at other compounds, ex calcium sulfate, which may have a lesser ability to suppress the melting point of ice.  The from there is is how to make CaSO4 from CaCl2....

[mechanick]

But for a reaction that would create CaSO₄, like:
CaCl_2(aq) + MgSO_4(aq)  CaSO_4(aq) + MgCl_2(aq)

Wouldn't that only depress the freezing point by turning the original three ions (Ca²⁺, Cl^-, Cl^-) into a solution with five ions present to prevent coagulation ([Ca²⁺, SO₄^2-, Mg²⁺, Cl^-, Cl^-)?

[408]

Not since CaSO4 is insoluble and will not depress the freezing point.  In the reaction you wrote only the MgCl2 is soluble, and as such having the ability to suppress the freezing point.

[AWK]

CaSO4 is weakly soluble (ca. 2g/liter)

[stewie griffin]

Also don't forget that colligative properties only depend on the number of moles involved, not the identity of the actual element/molecule/ion involved.
So when one mole of CaCl₂ dissolves we get 3 moles of ions (one of Ca⁺ and 2 of Cl^-). So in the freezing point depression equation we take into account those 3 moles.
However with something like a mole of CaSO₄ dissolves we only would get two moles of ions (assuming it completely dissolved in water). We'd get one mole of Ca⁺ ions and one mole of SO₄^2- ions. When we take into account that there's only 2 moles of ions here, the freezing point depression calculation will show that the freezing point is not lowered as much as in the case of calcium chloride.
So in other words, think about how to reduce the total number of moles of ions produced.