If 2.5 grams of calcium bromide reacted with excess lithium oxide, how many grams of bromide product would be formed?

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

i know insulin is primary but i dont know the others im sorry

explanation:

1. they vary with a higher number because boiling is way hotter than melting