C.J. Stroud has officially claimed the AFC Offensive Player of the Week award following a dominant performance against the Los Angeles Chargers. The Houston Texans quarterback earned the honor after leading his team to a decisive 44-22 victory in Week 10, capping a standout week with hardware from the NFL. Stroud’s return to Los Angeles was marked by immediate impact, as he orchestrated an eight-play, 73-yard opening drive that set the tone for the contest.
Texans QB ties career high with 384-yard, 125.5-rating masterpiece in 44-22 rout
Statistically, Stroud delivered a near-perfect outing, posting a 125.5 passer rating while completing 60.0% of his attempts for 384 yards. His five touchdown passes tied his career high and marked only the second time in his professional career he has thrown for over 350 yards while leading five scoring drives through the air. The Texans struck first on a five-of-five, 53-yard sequence that culminated in a touchdown pass to Xavier Hutchinson, establishing an early 7-0 lead that proved difficult for the Chargers to overcome.
Stroud’s performance placed him at the top of the league’s statistical leaders for the week, ranking first in both passing yards and touchdowns. He also recorded the third-highest passer rating in the NFL and finished in the top five for completion percentage. This milestone extends his impressive trajectory, following a similar multi-touchdown effort in Week 9 of the 2023 season against Tampa Bay, and solidifies his status as a premier offensive force in the American Football Conference.
Source: Yahoo Sports
