J.R. S. answered 12/14/20
Ph.D. University Professor with 10+ years Tutoring Experience
Combustion of butane:
2C4H10 + 13O2 ==> 8CO2 + 10H2O
oxygen atoms used in combustion = 13 moles O2 x 6.02x1023 molecules / mole x 2 atoms / molecule = # of oxygen atoms
oxygen atoms in product = 8 mol CO2 x 6.02x1023 molecules / mole x 2 atoms / molecule + 10 mol H2O x 6.02x1023 molecules / mole s 1 atom / molecule = # oxygen atoms in the product
Sarah L.
Combustion of Butane12/14/20