Answer:
Up to [tex]18\; \rm mol[/tex] [tex]\rm CO_2[/tex].
Explanation:
The equation for this reaction: [tex]2\; \rm C_3 H_7 OH + 9\; O_2 \to 6\; CO_2 + 8\; H_2O[/tex] is indeed balanced.
Notice that the coefficient of [tex]\rm C_3 H_7 OH[/tex] is [tex]2[/tex] while that of [tex]\rm CO_2[/tex] is [tex]6[/tex].
Therefore, each unit of this reaction would consume [tex]2[/tex] units of [tex]\rm C_3H_7OH[/tex] while producing [tex]6[/tex] units of [tex]\rm CO_2[/tex]. In other words, this reaction would produce three times as much [tex]\rm CO_2\![/tex] as it consumes [tex]\rm C_3H_7OH\![/tex].
If all that [tex]6\; \rm mol[/tex] of [tex]\rm C_3H_7OH[/tex] molecules are actually consumed through this reaction, three times as many [tex]\rm CO_2[/tex] molecules would be produced. That corresponds to [tex]3 \times 6\; \rm mol = 18\; \rm mol[/tex] of [tex]\rm CO_2\![/tex] molecules.
