41.4 grams of oxygen is essentially required for complete combustion of 3 moles of butane gas.
<h3>What is combustion?</h3>- Combustion is a chemical reaction that often involves the presence of oxygen and produces heat and light in the form of flames.
- Because of the nature of the chemical reaction and the fact that more energy is produced than can be released into the surrounding medium, the rate or speed at which the reactants combine is high.
- As a result, the temperature of the reactants is increased, accelerating the process even further.
As you know, combustion reaction is a chemical reaction in which a substance reacts with the atmospheric oxygen (O₂) to produce carbon dioxide and water. Therefore, balanced combustion reaction of butane (C₄H₁₀) will be as follows:
C₄H₁₀+O₂→4CO₂+5H₂O
You can observe in the above reaction that, moles of oxygen are required for combustion of 1 mole of butane.
Therefore, going by unitary method, for complete combustion of 3 moles of butane, (×0.398) moles of oxygen would be required.
Now, we know, number of moles =
Molar mass of oxygen is 16 g/mol
We need to find a given mass which is the mass of oxygen required when the number of moles is equal to 2.59.
Thus, given mass of oxygen or mass of oxygen required = number of moles of oxygen × molar mass of oxygen
Mass of oxygen required = 2.59 moles × 16 g/mol = 41.4 g.
Hence, 41.4 grams of oxygen is essentially required for complete combustion of 3 moles of butane gas.
To learn more about combustion, refer to
#SPJ4