Thermochemical Equations 07a (kJ) Consider the following thermochemical equation

Question

Thermochemical Equations 07a (kJ)

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

Part A

Calculate the heat associated with the consumption of 1.158 mol of O2 in this reaction.

Use the correct sign for q

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Part B

Calculate the heat associated with combustion of 29.46 g of butane.

Use the correct sign for q

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Part C

Calculate the mass of butane that must be burned in order to heat 33.09 kg of water from 22.31 C to 85.05 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.

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Explanation / Answer

A)

From the reaction:

15 mol of O2 = -5314.6 kJ;

then 1.158 mol of O2 = 1.158/15*-5314.6 = 410.28712 kJ released

b)

mol of butane = mass/MW 29.46/58.12 = 0.50688 mol of butane

2 mol of butane = -5314.6 kJ;

Q = 0.50688/2*5314.6 = 1,346.93 kJ released

c)

First, calculate heat required

Q = m*C*(Tf-Ti) = 33.09*4.184*(85.05-22.31)= 8,686.262 kJ

so

Q = -8,686.262

Q/HRxn = n

n = 8686.262 / (5314.6/2) = 3.268 moles of butane

mass = mol*MW = 3.268*58 = 189.544 g of butane

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