Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . suppose 25.1 g of octane is mixed with 39. g of oxygen. calculate the maximum mass of water that could be produced by the chemical reaction. round your answer to significant digits.

30.47 g H2O produced.

Explanation:

molecular weights:

mwOctane = 114 g/mol

mwO2= 32 g/mol

mwCO2 = 44 g/mol

mwH2O= 18 g/mol

Balancing the equation:

2 CH3(CH2)6CH3 + 25 O2 = 16 CO2 + 18 H2O

Present moles:

gOCtane= 25.1g

gO2= 39g

molOctane = gOCtane/mwOctane => 0.2202 mol

molO2 = gO2/mwO2 => 1.2188 mol

check ratios based on chemical equation:

Octaneratio = molOctane/2mol  => 0.1101

O2ratio = molO2/25mol => 0.0488

since Oxygen ratio is smaller, the this is the limit reactant:

25mol of O2 produces 18 moles of Water, then 1.2188 moles of  oxygen produces:

(25mol/18mol)*molO2 = 1.69 mol

of waters produced.

1.6927 mol*mwH2O = 30.47 g

H2O produced.