c3H8 + 5O2 = 3CO2 + 4H2O When 44.0 grams of propane (C3H8) under goes complete combustion, how many grams of water will be produced?
Added by Travis H.
Step 1
- Carbon (C): 12.01 g/mol - Hydrogen (H): 1.01 g/mol Molar mass of C\(_3\)H\(_8\) = 3(12.01 g/mol) + 8(1.01 g/mol) = 36.03 g/mol + 8.08 g/mol = 44.11 g/mol Show more…
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