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Ответ:

rodrickahammonds

01.08.2020

Chemistry

D) Methane CH4

Explanation:

Hello,

In this case, among the options, assuming for all the fuels a mass of 100 g, the combustion reaction reaction and stoichiometry for carbon dioxide is shown below:

A) Benzene C6H6

$C_6H_6+\frac{15}{2} O_2\rightarrow 6CO_2+3H_2O$

$m_{CO_2}=100gC_6H_6*\frac{1molC_6H_6}{78gC_6H_6}*\frac{6molCO_2}{1molC_6H_6} *\frac{44gCO_2}{1molCO_2} =347.4gCO_2$

B) Cyclohexane C6H12

$C_6H_{12}+9 O_2\rightarrow 6CO_2+6H_2O$

$m_{CO_2}=100gC_6H_{12}*\frac{1molC_6H_{12}}{84gC_6H_{12}}*\frac{6molCO_2}{1molC_6H_{12}} *\frac{44gCO_2}{1molCO_2} =314.3gCO_2$

C) Glucose C6H12O6

$C_6H_{12}O_6+6 O_2\rightarrow 6CO_2+6H_2O$

$m_{CO_2}=100gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molCO_2}{1molC_6H_{12}O_6} *\frac{44gCO_2}{1molCO_2} =146.7gCO_2$

D) Methane CH4

$CH_4+2O_2\rightarrow CO_2+2H_2O$

$m_{CO_2}=100gCH_4*\frac{1molCH_4}{16gCH_4}*\frac{6molCO_2}{1molC_6H_{12}} *\frac{44gCO_2}{1molCO_2} =1650gCO_2$

Therefore, for equal masses D) Methane CH4 yield the greatest mass of carbon dioxide.

Regards.

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