How many grams zinc phosphate (Zn3(PO4)2, M. M.=386.11 g/mol) will be produced if 835.0 g of zinc (Zn, M. M.=65.38 g/mol) reacts completely in the following equation: 3 Zn + 2 H3(PO4) → 3 H2 + Zn3(PO4)2

9112 g Zn3(PO4)2
1644 g Zn3(PO4)2
4931 g Zn3(PO4)2
4556 g Zn3(PO4)2

Mgo2 is the oxidation state of manganese oxide

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.