Answer:
3.51 g of oxygen per gram of gasoline is required.
Step-by-step explanation:
Solution:
First of all we will write the balance chemical equation.
C8H18 + 12.5O2 → 8CO2 + 9H2O
This equation shows that,
1 mole of gasoline react with 12.5 mole of oxygen for complete burning.
mass of one mole of gasoline = 8×12 + 18×1 = 114 g
mass of 12.5 mole of oxygen = 12.5 (16×2) = 400 g
Formula:
mass of oxygen per gram of gasoline = (400 / 114) = 3.51
so, 3.51 g of oxygen require for per gram of gasoline.