Oxygen hydrogen carbon dioxide

# A Compound Containing 3 Atoms Of Carbon And 8 Atoms Of Hydrogen Is Combined In A Reaction With Oxygen Molecules. The Two End Products Of This Equation Are Carbon Dioxide And Water. What Element Should You Look At First In Balancing This Equation? A.oxygen B.hydrogen C.carbon D.dioxide

The schematic equation can be written as:

C3H8 + O2 = CO2 + H2O.

For balancing this equation, firstly we have to consider atoms of HYDROGEN available for the reaction.

There are 8 on the Left Hand Side(L.H.S).

Thus, we have to write 4 H2O on the Right Hand Side(R.H.S).

Thus the equation now becomes,

C3H8 + O2 = CO2 + 4H2O.

Next,we have to check the number of Carbon atoms available for reaction on the L.H.S.

There are 3. Thus, we have to write 3CO2 on the R.H.S.

Thus the equation now becomes,

C3H8 + O2 = 3CO2 + 4H2O.

lastly, we have to balance the Oxygen atoms.

Look at the total number of O-atoms on the R.H.S.

Totally, we can see 10 O-atoms on the R.H.S.

Thus, we write 5O2 on the L.H.S.

Thus the equation now becomes,

C3H8 + 5O2 = 3CO2 + 4H2O.

This is the final fully balanced equation.

C3H8 + O2 = CO2 + H2O.

For balancing this equation, firstly we have to consider atoms of HYDROGEN available for the reaction.

There are 8 on the Left Hand Side(L.H.S).

Thus, we have to write 4 H2O on the Right Hand Side(R.H.S).

Thus the equation now becomes,

C3H8 + O2 = CO2 + 4H2O.

Next,we have to check the number of Carbon atoms available for reaction on the L.H.S.

There are 3. Thus, we have to write 3CO2 on the R.H.S.

Thus the equation now becomes,

C3H8 + O2 = 3CO2 + 4H2O.

lastly, we have to balance the Oxygen atoms.

Look at the total number of O-atoms on the R.H.S.

Totally, we can see 10 O-atoms on the R.H.S.

Thus, we write 5O2 on the L.H.S.

Thus the equation now becomes,

C3H8 + 5O2 = 3CO2 + 4H2O.

This is the final fully balanced equation.