Question

Liquid octane (CH 3 (CH 2 ) 6 CH 3 ) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O) Suppose 8.0 g of is with 39.5 of oxygenCalculate the minimum mass of octane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits

Answer 1

Octane can be written as C8H18.

The balanced equation for combustion of octane is:

2C8H18 + 25 O2 ==> 16CO2 + 18H2O

Finding the limiting reactant:

10.g C8H18 x 1 mol C8H18 / 118 g = 0.0847 moles (÷2->0.042)

61.2 g O2 x 1 mol O2 / 32 g = 1.91 moles (÷25->0.077)

C8H18 is the limiting reactant and will dictate the maximum amount of H2O that can form.

Maximum H2O formed:

0.0847 moles C8H18 x 18 moles H2O / 2 moles C8H18 x 18 g H2O/mol = 14 g H2O formed (2 sig.figs.)

Step-by-step explanation:

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