How much air is needed to burn 23 grams of ethyl alcohol? How many moles of carbon monoxide

How much air is needed to burn 23 grams of ethyl alcohol? How many moles of carbon monoxide (IV) and water does this get?

From the air components, oxygen will participate in the combustion reaction, from the reference data it is known that the volume fraction of oxygen in the air is φ = 21%, thus the volume of all air is Vair = V (O2) / φ, where V (O2) is the volume of oxygen that has entered into reaction with ethanol. V (O2) = n (O2) * Vm, where n (O2) is the amount of oxygen substance, Vm = 22.4 l / mol is the molar volume
Let’s write down the reaction scheme and arrange the coefficients:
2С2H5OH + 7O2 = 4CO2 + 6H2O
Let us determine the amount of ethanol substance n (et) = m (et) / M (et), where m (et) is the mass of ethanol, M (et) = 46 g / mol is the molar mass of ethanol.
n (et) = 23/46 = 0.5 mol
From the reaction equation it can be seen that 7 mol of oxygen is needed for 2 mol of ethanol, then:
n (O2) = n (et) * 7/2 = 0.5 * 7/2 = 1.75 mol
V (O2) = 1.75 * 22.4 = 39.2 L
Vair = 39.2 / 0.21 = 186.7 l