Write the balanced combustion equation for solid barium. Remember that combustion is a type of synthesis reaction.


2Ba + O2 ➡ 2BaO


Ba + O ➡ BaO


2Ba + O ➡ Ba2O


2Ba + O2 ➡ Ba2O2

Verified Answer :

2Ba + O2 --> 2BaO
(due to Ba^2+ & O^2- coming together).

Answer:

2Ba + O2 --> BaO

Explanation:

A combustion reaction is a reaction between a fuel with oxygen. Is this case, the fuel is Barium, which is a metal of group 2A, so it has 2 electrons in its valence shell. It must donate these 2 electrons to be stable and will be the cation \(Ba^{+2}\).

Oxygen is in group 6A, so it has 6 electrons in its valence shell, so must gain 2 electrons to be stable, and will be the anion \(O^{-2}\). So, they will form an ionic compound, and we need to change their charges without the signal to form its formula:

Ba2O2 = BaO

The reaction is:

Ba + O2 --> BaO

Balancing:

2Ba + O2 --> 2BaO