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
