I’m guessing the major error is that the -0.8V is “versus reversible hydrogen electrode ”, so it’s a half-reaction and you need to fill in the correct other half reaction, and there is probably H2 involved. Then you would dig out some free energy values to see how efficient it is.
(I only read the abstract.)
Fixed the math using chat gpt :P.
Reaction: 3CO2 + 12H2 -> C3H8 + 6O2
Number of moles of electrons (n): n = 3 * 4 = 12
Total charge (Q) in Coulombs: Q = (-395 * 1 * 100 * 3600 * 0.91 / 1000) * 12 * 96485
Voltage (V): V = -0.8
Total energy for given current density: E = Q * V
Moles of CO2 in 1 ton: n_CO2 = (1000 * 1000) / 44.01
Total charge for 1 ton of CO2: Q_total = (12 / 3) * n_CO2 * 96485
Adjusted charge considering Faradaic efficiency: Q_adjusted = Q_total * 0.91
Total energy for 1 ton of CO2: E_total = Q_adjusted * V
Convert to kWh: E_kWh = E_total * 2.778e-7 E_kWh ≈ 17707.4 kWh
Gives a specific CO2 emission for propane of 13.8 kg carbon / kWh fuel, so your numbers are at least vaguely credible. But your reaction is devoid of CO2, so talking about tons of CO2 is still odd.
6 O on the left, 12 on the right.
24 H on the left, 8 on the right.