In the reaction 2AgI + Na2S → Ag2S +2NaI, how many moles Ag2S will be formed if 85.0 g Na2S is used? How many grams of As are needed to form 0.380 moles Na3AsO3 in the reaction 2As + 6NaOH → 2Na3AsO3 + 3H2?

8.5gNaI × 1molNaI/78gNaI × 234.7g AgI / 1 MOL AgI × 2Mol Ag I / 1mol Na2S=51.2 g of AgI reacts with Na2 S

How many grams NaOH will be needed to form 0.380 moles Na3AsO3 in the reaction 2As + 6NaOH → 2Na3AsO3 + 3H2?

In the reaction 2AgI + Na2S → Ag2S +2NaI, how many moles Ag2S will be formed if 85.0 g Na2S is used?

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.

Go ask spider man