Question

# The propane fuel (C3H8) used in gas barbeques burns according to the following thermochemical equation: C3H8(g)+5O2(g)...

The propane fuel (C3H8) used in gas barbeques burns according to the following thermochemical equation:
C3H8(g)+5O2(g)?3CO2(g)+4H2O(g)

If a pork roast must absorb 1700kJ to fully cook, and if only 14% of the heat produced by the barbeque is actually absorbed by the roast, what mass of CO2 is emitted into the atmosphere during the grilling of the pork roast?

Mass of CO2 emitted = 680 g        (with two significant figures)

Explanation

Heat absorbed by roast = 1600 kJ

Heat produced by combustion = (Heat absorbed by roast) / (percentage of heat absorbed)

Heat produced by combustion = (1600 kJ) / (14 %)

Heat produced by combustion = (1600 kJ) / (14 / 100)

Heat produced by combustion = (1600 kJ) / (0.14)

Heat produced by combustion = 11428.6 kJ

Moles of propane burned = (Heat produced by combustion) / ($\dpi{100} \small \Delta$H)

Moles of propane burned = (11428.6 kJ) / (2217 kJ/mol)

Moles of propane burned = 5.155 mol

moles CO2 produced = 3 * (Moles of propane burned)

moles CO2 produced = 3 * (5.155 mol)

moles CO2 produced = 15.465 mol

mass CO2 produced = (moles CO2 produced) * (molar mass CO2)

mass CO2 produced = (15.465 mol) * (44.01 g/mol)

mass CO2 produced = 680.6 g

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