Question

How much propane, C5H12 (l), in both g and mL would you need to combust in order to produce enough heat to heat up 2 cups of water, or approximately 5.00 x 102 mL, from 22.0 ̊C to 95.0 ̊C? Assume that all the heat from the combustion directly goes to heating the water, no heat is lost to the surroundings. (density of water is 1.00 g/mL and density of C5H12 is 0.624 g/mL; ΔH°f, pentane = 121.7kJ/mol; use your book or notes to look of the values you need to calculate the heat of the reaction; assume gaseous water is produced).

Answer 1

Hi,

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Given reaction is combustion reaction of pentane (C₅H₁₂):

C5H12 (1) + 802(g) +5C02(g) + 6H20

Enthalpy change of this reaction is given by,

ΔΗραη = ΣΔΗ? (reactants) –ΣΔΗ? (reactants)

:. ΔΗγκη = 5.ΔΗ (CO2(g) + 6.ΔΗ (Η2Ο)1-ΔΗ (C; Η12106) + 8.ΔΗ (O2(g)

:: Arm = [(5x -393.5 kJ/mol) + (6 x -241.8 kJ/mol-1-121.7 kJ/mol + (8 x 0 kJ/mol

.

:. AH = -3418.3 kJ/mol – 121.7 kJ/mol

:: AHren = -3296.6 kJ/mol

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This heat will be used to raise the temperature of water.

We have,

Change in temperature,

AT = 95°C -22°C = 73°C

Heat capacity of water, C=4.184 J/g^oC

Mass of water = density x volume,

m = 500 mL x1g/mL = 500 g

Heat required to raise the temperature of water from 22^oC to 95^oC is given by:

H = mCAT = 500 g X 4.184 J/gºC x 73°C

:: H = 152716 J = 152.716 kJ

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No.of moles of propane required to produce this much heat is given by,

H 152.716 kJ n= AH - 3296.6 kJ/mol

:in=0.04632 mol

We have, molar mass of propane, M= 44 g/mol

Therefore, mass of propane required:

m=nx M = 0.04632 mol x 44 g/mol

..m = 2.038 g

Volume of propane:

m density 2.038 g 0.624 g/mL

:: V = 3.267 ml

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