Answer:
[tex]m_{CO_2}=46.6gCO_2[/tex]
Explanation:
Hello,
In this case, considering the combustion of propane:
[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)\ \ \ \Delta _CH=-2220.0 kJ/mol[/tex]
We can compute the burnt moles of propane as shown below:
[tex]n=\frac{-784kJ}{-2220.0 kJ/mol} =0.353molC_3H_8[/tex]
Then, by noticing propane and carbon dioxide are in a 1:3 molar ratio, we can compute the grams carbon dioxide by using the shown below stoichiometric procedure:
[tex]m_{CO_2}=0.353molC_3H_8*\frac{3molCO_2}{1molC_3H_8} *\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=46.6gCO_2[/tex]
Best regards.
