The combustion of propane is illustrated below. C3H3(g) + 502(9) --> 3CO2(g) + 4H2O(1) AH = -2550 kJ/mol.
1. If 18.0 grams of propane (44.11 g/mol) burns, how much energy will be produced?
figs)

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ava4460

10.11.2021

Chemistry

omg

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The combustion of propane is illustrated below. C3H3(g) + 502(9) --> 3CO2(g) + 4H2O(1) AH = -2550 kJ/mol. 1. If 18.0 grams of propane (44.11 g/mol) burns, how much energy will be produced? figs)