2NaHCO3 → Na2CO3 + H2CO3Determine the theoretical yields of each product using stoichiometry if the mass of the NaHCO3 sample is 3.55 grams. (Show work for both)In an actual decomposition of NaHCO3, the mass of one of the products was measured to be 2.16 grams. Identify which product this could be and justify your reasoning. Calculate the percent yield of the product identified in part B. (Show your work)

Crust,mantle,core,earth

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.