A 26.1 mL sample of a 1.24 M potassium chloride solution is mixed with 15.0 mL of a 0.890 M lead(II) nitrate solution and this precipitation reaction occurs: 2KCl(aq)+Pb(NO3)2(aq)→PbCl2(s)+2KNO 3(aq)

The solid PbCl2 is collected, dried, and found to have a mass of 2.47 g. Determine the limiting reactant, the theoretical yield, and the percent yield.

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.

They both go downwards and the can break apart rocks 

Seems cool i guess ttyl