Answered

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for the complete combustion of 47 g of gasoline (octane, C8H18) , the mass of oxygen consumed is

a)69.20g
b)82.45g
c)138.4g
d)164.9g

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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