How would you determine the limiting reactant if given 50.00 ml of 0.250 M Pb(NO3)2 reacting with 25.00 ml of 0.350 M CaCl2?

Question

Natalie Anton · Accepted Answer

#CaCl_2(aq) + Pb(NO_3)_2(aq) rarr Ca(NO_3)_2(aq) + PbCl_2(s)darr#

The stoichiometric equation is given above. But chemistry is an experimental science. You simply have to know that all nitrates are soluble in water, and most halides are soluble in water, EXCEPT for silver chloride, lead chloride, and mercurous chloride (#Hg_2Cl_2#).

Since lead chloride is insoluble, it will separate as a solid from solution.

I apologize for the spray, but before I started the solution, I had to say this.

We can finally determine the amounts of the metal reagents because the equation indicates that one equivalent of calcium chloride and one equivalent of lead nitrate will react.

Moles of lead nitrate: #50.00xx10^(-3)cancelLxx0.250*mol*cancel(L^(-1))# #=# #??mol#.

Moles of calcium chloride: #25.00xx10^(-3)cancelLxx0.350*mol*cancel(L^(-1))# #=# #??mol#.

It is obvious that there will be too much lead ion, but the math will need to be done.

Jose Ayala · Answer

undefined Determine moles of each reactant using given volume and molarity. Find the mole ratio between reactants using the balanced chemical equation. Identify the reactant with the smaller mole ratio as the limiting reactant.