Question

I am posting this problem because I've had trouble following the steps listed in similar problems. Please don't answer if you are only going to copy/paste a version of those same steps; It will not be helpful.

When a suspected drunk driver blows 186 mL of his breath through this breathalyzer, the breathalyzer produces an average of 323 mA of current for 10 s. Assuming a pressure of 1.0 atm and a temperature of 24 ∘C, what percent (by volume) of the driver's breath is ethanol?

Answer 1

we have the next reacction 2CH₃CH₂OH + O₂ -> 2CH₃CHO^- + 2OH^-

actually it is a redox reaction and we have than  CH₃CH₂OH -> CH₃CHOO^- + 2H⁺ + e

and the other reacction is 2e + O₂-> 2O^-2 if we sum we have

e + CH₃CH₂OH + O₂ -> CH₃CHOO^- + H₂O

we have than need a electron for begining the reacction of oxidation of ethanol to carboxilic acid

ok as 1mol = 96500columb and the A = columb/ s = A.S = columb =) 323x10^-3A.10s = 3.23columb =) the number of moles present is 3.23/96500 = 3.34x10^-5mol of ethanol now we know than

PV = nRT in equation for ideal gas where n = total of moles present in this volume of gas

P = 1atm V = 0.186l R = 0.082atm l /(mol K ) and T = 24 + 273 =297K =) n = PV/RT n = (1atm.0.186l)/( 0.082atm l /(mol K ).297K) = 7.63x10^-3 mol present in the gas like M is a propierty extensive is equal for any volume and the realtion % is equal

(n of ethanol/ n totals ) 100 = % of ethanol present =) (3.34x10^-5mol /7.63x10^-3 mol)100 = 0.44%