Determine the amount of oxygen required for combustion of 1.1 grams of butanol.

According to the coefficients of this equation, 6 oxygen molecules are required for the oxidation of 1 butanol molecule. This synthesizes 4 molecules of carbon dioxide.

Let’s calculate the amount of butanol available.

To do this, divide the weight of the available gas by the weight of 1 mole of this gas.

M C4H9OH = 12 x 4 + 10 + 16 = 74 grams / mol;

N C4H9OH = 1.1 / 74 = 0.015 mol;

The amount of oxygen will be.

N O2 = 0.015 x 6 = 0.09 mol;

Let’s calculate the gas volume. To do this, multiply the amount of substance by the volume of 1 mole of gas.

Its volume will be: V O2 = 0.09 x 22.4 = 2.016 liters;