Answer: 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:
[tex]C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O[/tex]
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 = [tex]\frac{1}{3}\times 1.42=0.473mol[/tex] 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 = [tex]\frac{2}{3}\times 1.42=0.946mol[/tex] 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 [tex]=\frac{22.4}{1}\times 0.946=21.2L[/tex] of volume
Hence, the volume of carbon dioxide measured at STP will be 21.2 L
