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) Δrxn= 2217kJ

Part A

If a pork roast must absorb 1500kJ to fully cook, and if only 12% 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. m=_____________________g

For solving this, first we have to calculate the amount of energy absorbed by the roast as follows:

2217 kJ * 0.12 = 266.04 kJ

1500kJ/266.04kJ = 5.6382 times the reaction has to happen

3 moles of CO2 / time * (5.6382 times) = 16.9147 moles of CO2

16.9147 moles of CO2 (44 grams/mol of CO2) = 744.25 g = 74.4 x 101 grams of CO2

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) / ( 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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