If 4.00 moles of gasoline are burned, what volume of oxygen is needed if the pressure is 0.953 atm, and the temperature is 35.0°C?

1 Answer
anor277
Jul 4, 2017

Well we will treat gasoline as octanes......#C_8H_18#, and we get a volume of

Explanation:

We need (i) a stoichiometric equation........

#C_8H_18(l) +25/2O_2(g) rarr 8CO_2(g) + 9H_2O(l)#

And so we need #25/2# equiv dioxygen gas.......

And thus if #4*mol# of octane are combusted we need #25/2xx4=50*mol# #"dioxygen gas."#

And (ii) given #V=(nRT)/P=(50*molxx0.0821*(L*atm)/(K*mol)xx308*K)/(0.953*atm)#

Approx....#1300*L# or #1.3*m^3# #"dioxygen gas"#.

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