Outside on a cold winter day a student opened up a cigarette lighter and poured 1.50 grams of liquid butane (b.p. -0.5 C°) into an otherwise empty balloon. The balloon was then brought indoors where the temperature was 22° C. If you assume that the pressure was 1 atmosphere, what would be the size of the balloon in liters?
If the balloon were to see a spark, it might explode. What volume of air would be needed for the combustion reaction to go to completion?

Question

Outside on a cold winter day a student opened up a cigarette lighter and poured 1.50 grams of liquid butane (b.p. -0.5 C°) into an otherwise empty balloon. The balloon was then brought indoors where the temperature was 22° C. If you assume that the pressure was 1 atmosphere, what would be the size of the balloon in liters?
If the balloon were to see a spark, it might explode. What volume of air would be needed for the combustion reaction to go to completion?

anonymous · Answer

pV= nRT 
n=m/M
V=nRT/p
V= mRT/pM
that is the volume but be careful to recalculate units to correspond to R = 8,314 JK-1mol-1
now when butane needs oxygen to burn and we know that there is 21% of oxygen in the air so when you calculate the volume of butane write the reaction of explosion and balance it, after that you calculate what volume of pure oxygen is needed for reaction and then just recalculate the volume bearing in mind that there is 21% of oxygen in air ;)

anonymous · Answer

can you please show me the steps in doing it . with numbers because I'm totally lost