38.55L
Given the following parameters
Mass of propane = 25.3 grams
Determine the moles of propane
Moles of propane = Mass/molar mass
Moles of propane = 25.3/44.1
Moles of propane = 0.574moles
The combustion of propane is given according to the equation below:
[tex]C_3H_8+5O_2\rightarrow3CO_2+4H_2O[/tex]According to stoichiometry, 1mole of propane produces 3 moles of CO2, the moles of CO2 required will be:
[tex]\begin{gathered} mole\text{ of CO}_2=3\times0.574 \\ moles\text{ of CO}_2=1.721moles \end{gathered}[/tex]Since 1 mole = 22.4L, volume of CO2 gas that will be produced is:
[tex]\begin{gathered} volume\text{ of CO}_2=1.721\times22.4 \\ volume\text{ of CO}_2=38.55L \end{gathered}[/tex]Hence the volume of carbon dioxide produced is 38.55L