Question

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?

Express your answer using two significant figures.

Answer 1

Answer 2

Answer 3

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