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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.

Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water Suppose 251 g of octane is mixed with 39 g of oxygen Calculate class=

Respuesta :

Answer:

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.