Answer:
[tex]V=70.4L[/tex]
Explanation:
Hello,
In this case, butane's combustion is represented by the following chemical reaction:
[tex]C_4H_{10}+\frac{13}{2} O_2\rightarrow 4CO_2+5H_2O[/tex]
Which is typically carried out in gas phase. Thus, for 45.6 g of butane, we employ its 1:4 molar relationship in the chemical reaction to compute the yielded moles of carbon dioxide by stoichiometry factors:
[tex]n_{CO_2}=45.6gC_4H_{10}*\frac{1molC_4H_{10}}{58gC_4H_{10}}*\frac{4molCO_2}{1molC_4H_{10}} =3.14molCO_2[/tex]
Next, by using the ideal gas equation at STP (273 K and 1 atm), we compute the produced liters of carbon dioxide as shown below:
[tex]PV=nRT\\\\V=\frac{nRT}{P}=\frac{3.14mol*0.082\frac{atm*L}{mol*K}*273K}{1atm} \\\\V=70.4L[/tex]
Best regards.
