How many grams of oxygen are required to rect with 10.0 grams of octane in the combustion of octane gasoline?

The stoichiometric reaction of the combustion of octane is C7H16 + 11O2 = 7 CO2 + 8H2O. Hence, for each mole of octane is equivalent to 11 moles of oxygen. 10.0 grams of octane divided by the molar mass (100g/mol) is equal to 0.1 mol. We multiply by 11 and the molar mass of oxygen (32g/mol), the answer is 35.2 grams oxygen.

In inorganic chemistry, a ternary compound is a compound containing three different elements. An example is sodium phosphate, Na3PO4. The sodium ion has a charge of 1+ and the phosphate ion has a charge of 3-. Therefore, three sodium ions are needed to balance the charge of one phosphate ion

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