Answer:
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
[tex]C_6H_6+\frac{15}{2} O_2\rightarrow 6CO_2+3H_2O[/tex]
[tex]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[/tex]
B) Cyclohexane C6H12
[tex]C_6H_{12}+9 O_2\rightarrow 6CO_2+6H_2O[/tex]
[tex]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[/tex]
C) Glucose C6H12O6
[tex]C_6H_{12}O_6+6 O_2\rightarrow 6CO_2+6H_2O[/tex]
[tex]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[/tex]
D) Methane CH4
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
[tex]m_{CO_2}=100gCH_4*\frac{1molCH_4}{16gCH_4}*\frac{6molCO_2}{1molC_6H_{12}} *\frac{44gCO_2}{1molCO_2} =1650gCO_2[/tex]
Therefore, for equal masses D) Methane CH4 yield the greatest mass of carbon dioxide.
Regards.
