Calculate the heat associated with the consumption of 1.542 mol of O2 in this re

Question

Calculate the heat associated with the consumption of 1.542 mol of O2 in this reaction.

Calculate the heat associated with combustion of 25.08 g of butane.

Calculate the mass of butane that must be burned in order to heat 23.83 kg of water from 22.31 C to 65.71 C. Assume no loss of heat in the transfer from the reaction to the water. The specific heat of water is 4.184 J/gC.

Consider the following thermochemical equation for the combustion of butane.
2C4H10(g)+15O2(g)8CO2(g)+10H2O(g)Hrxn=5314.6kJ

Explanation / Answer

2C4H10(g)+15O2(g) 8CO2(g)+10H2O(g) Hrxn = 5314.6kJ

a) 15 mol O2 = -5314.6 kj

so that,

1.542 mol O2 on consumption = -5314*1.542/15 = -546.3 kj

b)

No of mol of butane = 25.08 / 58 = 0.432 mol

so that

amount of energy released = 0.432*5314/2 = 1147.824 kj

c)

heat absorbed by water = m*s*DT

    = 23.83*10^3 *4.18*(65.71-22.31)

    = 4323.05 kj

2 mol butane = - 5314.6 kj

No of mol of butane = 4323.05*2/5314 = 1.63 mol

mass of butane = 1.63*58 = 94.54 grams

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