To determine how many moles of oxygen (O[tex]\(_2\)[/tex]) are needed to consume 0.412 moles of propene (C[tex]\(_3\)[/tex]H[tex]\(_6\)[/tex]), we need to use the balanced chemical equation provided:
[tex]\[ 2 \, \textrm{C}_3\textrm{H}_6(g) + 9 \, \textrm{O}_2(g) \rightarrow 6 \, \textrm{CO}_2(g) + 6 \, \textrm{H}_2\textrm{O}(g) \][/tex]
From this equation, we see that 2 moles of C[tex]\(_3\)[/tex]H[tex]\(_6\)[/tex] react with 9 moles of O[tex]\(_2\)[/tex]. This gives us a stoichiometric ratio of 2:9 between C[tex]\(_3\)[/tex]H[tex]\(_6\)[/tex] and O[tex]\(_2\)[/tex].
To find out how many moles of O[tex]\(_2\)[/tex] are needed for 0.412 moles of C[tex]\(_3\)[/tex]H[tex]\(_6\)[/tex], we use the stoichiometric ratio:
[tex]\[ \frac{9 \, \text{moles of O}_2}{2 \, \text{moles of C}_3\text{H}_6} \][/tex]
We multiply the number of moles of C[tex]\(_3\)[/tex]H[tex]\(_6\)[/tex] by this ratio:
[tex]\[ \text{Moles of O}_2 \, \text{needed} = 0.412 \, \text{moles of C}_3\text{H}_6 \times \frac{9 \, \text{moles of O}_2}{2 \, \text{moles of C}_3\text{H}_6} \][/tex]
Simplifying the expression:
[tex]\[ \text{Moles of O}_2 \, \text{needed} = 0.412 \times 4.5 \][/tex]
[tex]\[ \text{Moles of O}_2 \, \text{needed} = 1.8539999999999999 \][/tex]
So, to consume 0.412 moles of C[tex]\(_3\)[/tex]H[tex]\(_6\)[/tex], you need approximately 1.854 moles of [tex]\(\text{O}_2\)[/tex].
