Question

If you combust 45.6 g of butane (C4H10), how many L of CO2 are produced, at STP?

Answer 1

70

Step-by-step explanation:

Answer 2

`V=70.4L`

Step-by-step explanation:

Hello,

In this case, butane's combustion is represented by the following chemical reaction:

`C_4H_(10)+(13)/(2) O_2\rightarrow 4CO_2+5H_2O`

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:

`n_(CO_2)=45.6gC_4H_(10)*(1molC_4H_(10))/(58gC_4H_(10))*(4molCO_2)/(1molC_4H_(10)) =3.14molCO_2`

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:

`PV=nRT\\\\V=(nRT)/(P)=(3.14mol*0.082(atm*L)/(mol*K)*273K)/(1atm) \\\\V=70.4L`

Best regards.