What volume of carbon dioxide measured at stp will be formed by the reaction of 1.42 mol of oxygen with 0.900 mol of ethyl alcohol?

The volume of carbon dioxide measured at STP will be 21.2 L

Explanation:

We are given:

Moles of oxygen = 1.42 moles

Moles of ethyl alcohol = 0.900 moles

The chemical equation for the combustion of ethyl alcohol follows:

By Stoichiometry of the reaction:

3 moles of oxygen gas reacts with 1 mole of ethyl alcohol

So, 1.42 moles of oxygen gas reacts with = of ethyl alcohol

As, given amount of ethyl alcohol is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of oxygen gas produces 2 moles of carbon dioxide

So, 1.42 moles of oxygen gas will produce = of carbon dioxide

At STP:

1 mole of a gas occupies 22.4 L of volume

So, 0.946 moles of oxygen gas will occupy of volume

Hence, the volume of carbon dioxide measured at STP will be 21.2 L