41.4 grams of oxygen is essentially required for complete combustion of 3 moles of butane gas.
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+[tex]\frac{13}{2}[/tex]O₂→4CO₂+5H₂O
You can observe in the above reaction that, [tex]\frac{13}{2}[/tex] 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, ([tex]\frac{13}{2}[/tex]×0.398) moles of oxygen would be required.
Now, we know, number of moles = [tex]\frac{Given mass}{Molar mass}[/tex]
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
https://brainly.com/question/13251946
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