A balloon is filled with 0.896 g of butane gas, C 4 H 10 (g), and 4.553 g of oxy

Question

A balloon is filled with 0.896 g of butane gas, C4H10(g), and 4.553 g of oxygen and sealed. If the butane was combusted to form carbon dioxide gas and water vapor (yes, consider the water as a gas) the heat released would increase the temperature of the gases in the balloon. If the balloon was initially at 1.089 atm and 16.1 °C, what is the final volume of the balloon? Hint: You will need to determine the temperature of the gases after combustion and the specific heat of the gases, sgases, is 1.526 J/(g °C). Report your answer in liters to one decimal place. Show all work and do not round for the WHOLE question.

Explanation / Answer

heat released = heat gained by gases

1C4H10(g) + 13/2O2(g) ---> 4co2(g) + 5H2O(g)

DH0rxn = -2673.5kj/mol.

no of mol of butane = 0.896/58 = 0.01545 mol

no of mol of O2 = 4.553/32 = 0.1423 mol

limiting reactant -butane

excess O2 = 0.1423 - (0.01545*13/2) = 0.042 mol

heat released = -2673.5*0.01545 = -41.3 kj

heat gained by gases = m*s*DT

                   41.3*10^3   = (0.896+4.553)*1.526*(x-16.1)

                 x = final temperature of gas in ballon = 4983 c

V1 = initial volume of ballon = n*RT/P

initial volume of ballon = (0.01545+0.1423)*0.0821*(16.1+273.15)/1.089

              v1 = 3.44 L

V2 = final volume

= (0.01545*4+0.01545*5+0.042)*0.0821*(4983+273.15)/1.089

               = 71.74 L

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