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Sagot :
Answer: Octane will be used completely.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of octane}=\frac{2.3g}{114g/mol}=0.0202moles[/tex]
[tex]\text{Moles of oxygen}=\frac{12.4g}{32g/mol}=0.388moles[/tex]
The balanced chemical reaction will be
[tex]2C_8H_{18}+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)[/tex]
According to stoichiometry :
2 moles of octane require = 25 moles of [tex]O_2[/tex]
Thus 0.0202 moles of octane will require=[tex]\frac{25}{2}\times 0.0202=0.2525moles[/tex] of [tex]O_2[/tex]
Thus octane is the limiting reagent as it limits the formation of product and [tex]O_2[/tex] is the excess reagent.
Thus octane will be used completely.
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