J.R. S. answered 10/30/20
Ph.D. University Professor with 10+ years Tutoring Experience
2C6H14 + 19O2 ==> 12CO2 + 14H2O ... balanced equation
b) Combustion reaction
c) 45.6 g O2 x 1 mol O2/32 g x 14 mol H2O / 19 mol H2O x 18 g H2O/mol H2O = 18.9 g H2O
d) 389.44 L O2 x 12 L CO2 / 19 L O2 = 245.96 L CO2
e) 87.10 mol CO2 x 2 mol C6H14 / 12 mol CO2 x 86 g C6H14/mol = 1248 g C6H14
f) 508.33 g C6H14 x 1 mol C6H14 / 86 g x 14 mol H2O / 2 mol C6H14 = 41.38 moles H2O