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A chemistry journal requires that all new research papers be peer reviewed by other scientists before they are published. This

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ElenaW [278]3 years ago

5 0

Validity is the answer

Sindrei [870]3 years ago

4 0

Answer:

I think its number 2

Explanation:

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- From octane: $m_{CO_2}=1.54x10^5gCO_2$

- From ethanol: $m_{CO_2}=9.57x10^4gCO_2$

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Hello,

At first, for the combustion of octane, the following chemical reaction is carried out:

$C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O$

Thus, the produced mass of carbon dioxide is:

$m_{CO_2}=5.00x10^4gC_8H_{18}*\frac{1molC_8H_{18}}{114gC_8H_{18}}*\frac{8molCO_2}{1molC_8H_{18}}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=1.54x10^5gCO_2$

Now, for ethanol:

$C_2H_6O+3O_2\rightarrow 2CO_2+3H_2O$

$m_{CO_2}=5.00x10^4gC_2H_6O*\frac{1molC_2H_6O}{46gC_2H_6O}*\frac{2molCO_2}{1molC_2H_6O}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=9.57x10^4gCO_2$

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