2C8H18 + 25O2 → 16CO2 + 18H2O If you burn 3.1 L of octane gas (C8H18) at 300 K and 0.68 atm, how many grams of water will you produce?

Answer 1

2C8H18 + 25O2 → 16CO2 + 18H2O ... balanced equation

-Use ideal gas law PV = nRT to find moles of octane

-Covert moles octane to moles of water formed using stoichiometry of balanced equation

-convert moles of water to grams of water using molar mass of water

-moles of octane present

PV = nRT

n = PV/RT = (0.68 atm)(3.1 L) / (0.0821 Latm/Kmol)(300K) = 0.0856 mols octane

-moles H2O formed:

0.0856 mols octane x 18 mols H2O / 2 mols octane = 0.770 mols H2O

-grams H2O formed:

0.770 mols H2O x 18 g / mol = 13.9 g H2O