When 0.105 mol propane, c3h8 is burned in an excess of oxygen, how many moles of oxygen are consumed? the products are carbon dioxide and water. you have to properly balance the equation before you can answer this question?

Propane on reaction with Oxygen produces CO₂ and water;

                       C₃H₈  +  10 O₂    →    3 CO₂  +  4 H₂O

According to Balance equation,

                          1 mole of C₂H₈ required   =   10 moles of O₂

Then,

              0.105 moles of C₂H₈ will require   =   X  moles of O₂

Solving for X,

                              X  =  ( 10 mol × 0.105 mol) ÷ 1 mol

                              X  =  1.05 mol of O₂

Result:
            1.05 mol of O₂ is required to consume 0.105 mol of C₂H₈ completely.

It has to be b i hope this