If a car produces 88 grams of co2 how many grams of o is used up in the reactionin c2h6o o2

B. (4.0 mol C2H6O) x (3/1) = 12 mol O2 

C. Supposing all the C in the CO2 came from ethanol: 
(88 g CO2) / (44.00964 g CO2/mol) x (3/2) x (31.99886 g O2/mol) = 96 g O2 

D. (125 g C2H6O) / (46.06867 g C2H6O/ mol) x (2/1) x (44.00964 g CO2/mol) = 
239 g CO2 

(125 g C2H6O) / (46.06867 g C2H6O/ mol) x (3/1) x (18.01532 g H2O/mol) =

Hbr dissociates to form h+ and br-ca(oh)2 dissociates to form ca2+ and 2oh-the h+ and oh- react to form waterassuming the reactants are dissolved in water, the net reaction is h+ + oh- —> h2oif you’re asked to show all components of the reaction, the complete reaction is 2hbr + ca(oh)2 —> cabr2 + 2h2o

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.