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Sagot :
The grams of oxygen that are required to react with 11.0 grams of octane is 38.6 grams.
The balanced chemical reaction can be depicted as follows:
C₈H₁₈ + 25/2 O₂ ------> 8CO₂ +9 H₂O
Atomic mass of C = 12 u
Atomic mass of H = 1 u
Atomic mass of O = 16 u
Molar mass of C₈H₁₈ = 8*12+18*1 = 96+18 = 114 g/mol .
Molar mass of O₂= 2*16 = 32 g/mol
Mass of C₈H₁₈ = 11 g
Moles = mass/molar mass
Moles of Octane = 11 g/ 114(g/mol) = 0.09649 mol .
From balanced chemical reaction :
1 mol Octane ------> 12.5 mol O₂
0.09649 mol octane -----> x
x = 0.09649 mol octane* 12.5 mol O₂/1 mol octane
x = 1.206 mol O₂
Moles of Oxygen = 1.206 mol
Mass of oxygen= moles of oxygen* molar mass of oxygen
Mass of oxygen= 1.206 mol * 32 (g/mol) = 38.59 g≅38.6 g
To learn more about moles check the link below:
https://brainly.com/question/15356425
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