a) Write a balanced chemical equation for the complete combustion of heptane. b)

Question

a) Write a balanced chemical equation for the complete combustion of heptane. b) Calculate the volume (in liters) of air at 34 degrees Celsius and 0.s atmosphere that is completely 1263 grams of heptane. Assume that air is 21.0 percent O2 by volum 02| 170.ly 367 e) When I mol of heptane is reacted with pure oxygen, O2,-4812 kJ or heat is released. Calculate heat of formation AH"r of heptane given that the Hof of H2O(l)is-285.3 kl mori and Hof of co 2(g) is-393.5 kJ mor. (Note: I expect you to calculate the 11 fofheptane from the data in this problem.) Dork VJ d) Assuming that all of the heat evolved in burning 12.63 grams of heptane is transferred to 8.17 kilograms of water (specific heat-4.184Jg deg C) initially at 25.5 degrees Celsius, calculate the increase in temperature of the water. 2 As 4.8 g of NO are added to an amount of O2, After the reaction occurs 1.67 moles of N203 are p to the equation, 4M0(g) + 02(g)--> 2N203(1) of the following questions. If a question can not he ancwond

Explanation / Answer

d)

Heat released in burning 1 mole of heptane = -4812 kJ (as per Part C)

Molar mass of Heptane (C7H16) = 12 * 7 + 16 * 1 = 100 gm/mol

Number of moles of Heptane in 12.63 grams of heptane = 12.63/100 = 0.1263

Heat released = number of moles * molar heat = 0.1263 * -4812 = -607.756 kJ

Heat evolved = mass * specific heat capacity of water * change in Temperature

607.756 * 10^3 = 8170 * 4.184 * (change in temperature)

Change in temperature = 17.7793236928

Hence the final temperature of water = Initial temperature + change in temperature = 25.5 + 17.779 = 43.279C

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