For the electrolysis described below:
what volume of pure oxygen gas is evolved at 16°C and 768 Torr for every kilogram of gold produced??
4Au(CN)4- + 12OH- 4Au + 16CN- + 3O2 + 6H2O
My attempt: (.00375*.0821*289.15)/1.01 ATM
= .08809 L but that is incorrect. Please explain.
Trevor H
Favorite Answer
From your equation I interpret that:
When 4 mol Au produced , then 3 mol O2 produced.
Molar mass Au = 196.97g/mol
1000g = 1000/196.97 = 5.08 mol Au produced
This will result in 5.08*3/4 = 3.81 mol O2 being produced.
Calculate volume using ideal gas equation:
PV = nRT
P = pressure = 768 torr = 1.0105 atm
V = What you want
n = mol = 3.81
R =- constant = 0.082057
T = temp = 16°C = 289K
Substitute
1.0105 * V = 3.81*0.082057*289
V = 90.352 / 1.0105
V = 89.4L
Volume of O2 produced = 89.4L
bumbray
Volume Of Oxygen