the balanced equation for the combustion of C₆H₁₄ is as follows
2C₆H₁₄ + 19O₂ ---> 12CO₂ + 14H₂O
stoichiometry of C₆H₁₄ to O₂ is 2:19
number of moles = mass present / molar mass
number of C₆H₁₄ moles reacted = 86.17 g / 86 g/mol = 1.00 mol
according to molar ratio of 2:19
2 mol of C₆H₁₄ reacts with 19 mol of O₂
then 1.00 mol of C₆H₁₄ reacts with - 19 / 2 x 1.00 = 9.5 mol of O₂
mass of O₂ required = 9.5 mol x 32 g/mol = 304 g
mass of O₂ required is 304 g
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