Amass of 13.56 g for an unknown fuel was combusted in a reaction vessel containing an unknown amount of oxygen. at the end of the reaction, there still remained 13.52 g of the fuel as well as 0.0654 g of water and 0.1198 g of carbon dioxide. the oxygen was completely consumed during the reaction. how many molecules of oxygen gas were initially present in the reaction vessel

The answer to your question is:  = 2.52 x 10²¹ molecules

Explanation:

mass of fuel = 13.56 - 13.53 = 0.04 g of fuel

mass of oxygen = ?

mass of water = 0.054 g

mass of carbon dioxide = 0.1198 g

To solve this problem just remember the Lavoisier law of conservation of mass that says: Mass is neither created nor destroyed in chemical reactions,

Reaction:

            Fuel   + Oxygen    ⇒    Water    +  Carbon dioxide

            0.04  +  mass O2  ⇒   0.054     +   0.1198

           mass O2 = 0.054 + 0.1198 - 0.04

           mass O2 = 0.1338 g

 MW O2 = 32

                        1 mol of O2   32 g

                           x                    0.1338g

                    x = 0.1338(1) / 32 = 0.0042 moles

                      1 mol 6,023 x 10 ²³  molecules

                    0.0042 moles     x

          x = 0.0042 (6,023 x 10 ²³ ) / 1 = 2.52 x 10²¹ molecules

- lead + phosphoric acid = hydrogen + lead(ii) phosphate

i don,t know so just answer mine