The complete combustion of propane, c3h8 (g), is represented by the equation: c3h8(g) + 5 o2(g) → 3 co2(g) + 4 h2o(l) δh° = -2220 kj how much heat is evolved in the complete combustion of 20.0 l propane at stp?

ΔH° =  -1815 kJ

Explanation:

The balanced chemical equation

                                C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)

tells us that 2220 kJ joules are released in the combustion of one mol propane,C3H8  . So what we need to solve this problem is to find how many moles of propane 20.0 L represent and do the calculation.

To do that, we will be using the Ideal Gas Law since we are told the volume,  temperature, and pressure.

PV = nRT ∴ n = PV/RT

P: 1 atm

V: 20.0 L

R= 0.08206 Latm/kmol ( R constant for ideal gases)

T= 25 ºC + 273 = 298 k  (Need to convert T to degree Kelvin)

Plugging the values

n = 1 atm x 20.0 L/ (0.08206 Latm/ k mol)

n = 0.82 mol

ΔH° =(-2220 kJ / 1 mol C3H8 ) 0.82 mol C3H8 = -1815 kJ

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