The following chemical equation 2C8H18(g)+ 25O2(g)→ 16CO2(g)+ 18H2O(g) is balance and 48.64 grams of oxygen is required in the combustion of octane in gasoline.
When burning of petrol happens and releases toxic hydrocarbons and carbon dioxide gasses with some amount of oxygen is known as combustion.
The balanced chemical equation is 2C8H18(g)+ 25O2(g)→ 16CO2(g)+ 18H2O(g),
number of moles = mass / molar mass
substituting the value in equation,
number of moles = 14 / 114g = 0.122 moles.
number of moles of oxygen 0.122 × 0.122 moles. = 1.52 moles.
1.52 = mass / 32
mass = 48.64 gram.
Therefore,48.64 grams of oxygen is required in the combustion of octane in gasoline.
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The balanced chemical equation would be as follows,
C8H18(g)+12.5 O2(g)→ 8 CO2(g)+ 9 H2O(g)
105.26 grams of oxygen are required to react with 14.0 grams of octane ( C8H18 ) in the combustion of octane in gasoline.
A chemical reaction is a process in which one or more substances, also known as reactants are converted to one or more different substances, known as products. A chemical reaction rearranges the constituent atoms of the reactants to create different substances as products
The balanced chemical equation would be as follows,\
C8H18(g)+12.5 O2(g)→ 8 CO2(g)+ 9 H2O(g)
114 grams of octane ( C8H18 ) require 400 grams of oxygen
1 gram of octane requires 400/114 grams of oxygen
30 grams of octane require 400/114×30 grams of oxygen
30 grams of octane require 105.26 grams of oxygen
Thus,105.26 grams of oxygen are required to react with 14.0 grams of octane ( C8H18 ) in the combustion of octane in gasoline.
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