Which conversion factor should you use first to calculate the number of grams of co2 produced by the reaction of 50.6 g of ch4 with o2? the complete combustion of methane is described by this equation: ch4(g) + 2o2(g) → co2(g) + 2h2o(l) 44.0 g co2 / 2 mol co2 16.0 g ch4 / 1 mol co4 1 mol ch4 / 16.0 g ch4 2 mol o2 / 1 mol co2

139.18 g.

Explanation:

For the balanced equation:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l),

It is clear that 1 mol of CH₄ react with 2 mol of O₂ to produce 1 mol of CO₂ and 2 mol of H₂O.

Firstly, we need to calculate the no. of moles of 50.6 g of CH₄:

no. of moles of CH₄ = mass/molar mass = (50.6 g)/(16.0 g/mol) = 3.1625 mol.

Using cross-multiplication:

1.0 mol of CH₄(g) produces → 1.0 mol of CO₂, from stichiometry.

∴ 3.1625 mol of CH₄(g) produces → 3.1625 mol of CO₂.

∴ The no. of grams of CO₂ produced = (no. of moles of CO₂)(molar mass of CO₂) = (3.1625 mol)(44.01 g/mol) = 139.18 g.

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.