C4H10(g) delta H(kj/mol)=-124.73 delta G=-12.71 delta S=310.0 calculate how much

Question

C4H10(g) delta H(kj/mol)=-124.73 delta G=-12.71 delta S=310.0

calculate how much heat is released when 0.72 grams of butane (C4H10) undergoes combustion at STE. Include the sign and report to 1 decimal place in kJ.

Using the same data as in the previous question, calculate how many moles of gas are released by the combustion of 43 grams of butane. Report to two decimal places.

Using the same data as in the previous questions, calculate how many liters of gas are released by the combustion of 45.04 grams of butane. Report to one decimal place and assume STP.

Using the same data as in the previous questions, calculate how many liters of gas are released by the combustion of 45.04 grams of butane. Report to one decimal place and assume STP.

Explanation / Answer

Solution :-

1) Balanced reaction equation

2 C4H10 + 13 O2 ----- > 8 CO2 + 10 H2O

Lets first calculate the enthalpy change of the reaction

Delta H rxn = sum of delta Hf product - sum of delta Hf reactant

                     = [(CO2*8)+(H2O*10)] –[C4H10*2]

                     = [(-393.5*8)+(-241.8*10)] –[-124.73*2]

                      = -5310.54 kJ

Now lets calculate the amount of heat given by the 0.72 g butane

(0.72 g butane * 1 mol /58.12 g )*(5310.54 kJ / 2 mol butane ) = 32.9 kJ

So the 0.72 g butane will give 32.9 kJ heat.

2) 2 mol butane gives total 8+10 = 18 mol gases

So lets calculate the moles of gases produced by 43 g butane

(43 g butane * 1 mol /58.12 g )* (18 mol gases / 2 mol butane ) = 6.66 mol gases

So it will form 6.66 mol gases

3) Now lets calculate the volume of the gases produced by 45.04 g butane

(45.04 g butane * 1 mol /58.12 g )* (18 mol gases / 2 mol butane )*(22.4 L / 1 mol ) = 156.2 L

So the volume of the gases produced at STP = 156.2 L

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