The amount, in grams, of [tex]CO_2[/tex] that would be produced from the burning of 3.0 mol amyl alcohol would be 660.15 grams
Stoichiometric calculation
From the equation of the reaction:
[tex]2C_5H_1_1OH + 15O2 ---- > 10CO_2 + 12H_2O[/tex]
The mole ratio of amyl alcohol to the [tex]CO_2[/tex] produced is 2:10.
Thus, for 3.0 mol of amyl alcohol, 15 mol of [tex]CO_2[/tex] would be produced.
Mass of 15 mol [tex]CO_2[/tex] = 15 x 44.01 = 660.15 grams
More on stoichiometric calculations can be found here: https://brainly.com/question/8062886
