Tide's Keelon Russell Named Shaun Alexander Award Player Of The Week

TUSCALOOSA, AL — Alabama quarterback Keelon Russell was named the Shaun Alexander Award Player of the Week after helping lead the Crimson Tide to a 50-36 win over Florida State.

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Russell accounted for 376 yards of total offense and four touchdowns in the victory.

The sophomore completed 23 of 30 passes for 320 yards and three touchdowns, with his scoring throws covering 21, 46 and 16 yards. All three touchdown passes came during a span of 2 minutes, 36 seconds before halftime.

Russell also made an impact on the ground, finishing with 56 rushing yards and a touchdown on three carries. His rushing score came on a 57-yard scramble.

The performance helped Alabama remain unbeaten at 3-0 on the season.

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