Answer: 2.17 g of bromide product would be formed
Explanation:
The reaction of calcium bromide with lithium oxide will be:

To calculate the moles :

As lithium oxide is in excess, calcium bromide is the limiting reagent.
According to stoichiometry :
1 mole of
produce = 2 moles of 
Thus 0.0125 moles of
will require=
of 
Mass of 
Thus 2.17 g of bromide product would be formed
CH3CH2CH2NH2+H2O<=>CH3CH2CH2NH3++OH-
Electrons would be the correct answer.
Answer:
2KClO3--------->2KCl + 3O2