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Sagot :
Answer: d) 164.9 g
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex] [tex]\text{Moles of octane}=\frac{47g}{114g/mol}=0.412moles[/tex]
The balanced chemical reaction is:
[tex]2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)[/tex]
According to stoichiometry :
2 moles of [tex]C_8H_{18}[/tex] require = 25 moles of [tex]O_2[/tex]
Thus 0.412 moles of [tex]C_8H_{18}[/tex] will require=[tex]\frac{25}{2}\times 0.412=5.15moles[/tex] of [tex]O_2[/tex]
Mass of [tex]O_2=moles\times {\text {Molar mass}}=5.15moles\times 32g/mol=164.9g[/tex]
Thus 164.9 g of oxygen is consumed.
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