If 0.647 mil of oxygen are used in the burning of propane, how many moles of CO2 are produced

1.733 x 10⁻⁴ mole CO₂(g)

Explanation:

Assuming the rxn given is at STP conditions ...

                        C₃H₈             +       5O₂      =>             3CO₂    +    4H₂O

Given:             Excess             =  0.647ml                ?(moles)            

                                          = 0.000647Liters

                                      = 6.47 X 10⁻⁴L/22.4L·mol⁻¹

                                        = 1.733 x 10⁻⁵ mol O₂

moles CO₂ = 3/5(2.89 x 10⁻⁴)mole CO₂ = 1.733 x 10⁻⁵ mole CO₂

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.