En un taller se plantea la siguiente ecuación química: I
CaCl2 + 2 NaBr
CaBr2 + 2 NaCl
(CaCl, 111 g/mol; NaBr 103 g/mol; Cabr, 200 g/mol; NaCl 58,5 g/mol)
Posteriormente se solicita que se indique el reactivo limitante cuando reaccionan 2229
de CaCl, con 309 g de NaBr.
En este caso la respuesta correcta para contestar el taller sería:
A CaCl, ya que sobran 103 g de este reactivo,
B. NaBr ya que hacen falta 103 g de este reactivo,
C CaCl, ya que hacen falta 103 g de este reactivo.
D NaBrya que sobran 103 q de este reactivo.

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.

Answer(2) electrons;