Balance the following reaction

Question

___Rb    +     ___RbNO₃     =>     ___Rb₂O    +     ___N₂

Answer 1

___Rb    +     ___RbNO₃     =>     ___Rb₂O    +     ___N₂

I would concentrate on balancing the most complex first. In this case, that is RbNO₃.

Put a 2 in front of RbNO₃ to balance the N₂.

_Rb + 2RbNO₃ => _Rb₂O + 1N₂

Doing that gives 6 Oxygen on the left, so put a 6 in front of the Rb₂O on the right to compensate.

_Rb + 2RbNO₃ => 6Rb₂O + 1N₂

There are 12 Rb on the right. 2 are covered in the 2RbNO₃ so we put a 10 in front of the elemental Rubidium to finish.

10Rb + 2RbNO₃ => 6Rb₂O +N₂