The air/fuel ratio that maximizes power has been shown to be about 12:1 or 12.5:1 and is about 20% richer than stoichiometric (14.7:1). 14.7:1 yields exhaust CO levels of less than 1%. 12:1 give about 6% CO. 0.5% is a target for emission control to address environmental concerns, not for engine power.
Since engine power is air-limited (not fuel-limited), for maximum power, you want to use up all of the air that gets ingested by the engine. So you richen up the mixture until the best tradeoff between over-rich power 'loss' and air utilization gives you maximum power. So up to a certain point a richer mixture burns slower, but produces more power. There’s a fuel/air ratio for optimum flame speed. Too lean, it slows; too rich it slows.
I refer you to:
Heywood, John B., Internal Combustion Engine Fundamentals, McGraw-Hill, 1988, p. 395
According to Heywood:
"Both ﬂame development and burning angles show a minimum for slightly rich mixtures ([phi is approximately] 1.2) and increase signiﬁcantly as the mixture becomes substantially leaner than stoichiometric. ... Faster burning engines (which have higher turbulence) are less sensitive to changes in mixture composition, pressure, and temperature than are slower burning engines (which have lower turbulence)."
In other words, richer (up to a point) burns faster (smaller burn angle), not slower. The minimum burn time appears to be right around the max power air/fuel ratio (phi about 1.2).
See pages 402 to 403, particularly Figure 9-25 at the bottom of page 403. Keep in mind that the figure is for a laminar flame, and which is not exact for the turbulent flame typical in an engine, but it is a clue and a start and not a bad guess for guessing the effect of changing parameters.
That said and to address your question, we find that generally peak power output comes at 4% CO.