Liquid octane ch3ch26ch3 reacts with gaseous oxygen gas o2 to produce gaseous carbon dioxide co2 and gaseous water h2o. if 26.7g of carbon dioxide is produced from the reaction of 10.3g of octane and 71.6g of oxygen gas, calculate the percent yield of carbon dioxide.

96.13% Yield Carbon

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= 10.3g

gO2= 26.7g

molOctane = gOCtane/mwOctane => 0.0904 mol

molO2 = gO2/mwO2 => 0.8344 mol

check ratios based on chemical equation:

Octaneratio = molOctane/2mol  => 0.0452

O2ratio = molO2/25mol => 0.0334

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/16mol)*molO2 => 1.3037 mol

of waters produced.

1.6927 mol*mwCO2 => 74.4788 g

should be produced, CO2 Produced,

Real Produced = 71.6 g

Ideally Produced = 74.48 g

Yield = Real Produced/Ideally Produced => 0.9613

96.13% Yield Carbon

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