Question

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

Answer 1

Answer: 2.17 g of bromide product would be formed

Step-by-step explanation:

The reaction of calcium bromide with lithium oxide will be:

`CaBr_2+Li_2O\rightarrow 2LiBr+CaO`

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of calcium bromide}=(2.5g)/(199.9g/mol)=0.0125moles`

As lithium oxide is in excess, calcium bromide is the limiting reagent.

According to stoichiometry :

1 mole of
`CaBr_2` produce = 2 moles of
`LiBr`

Thus 0.0125 moles of
`CaBr_2` will require=
`(2)/(1)* 0.0125=0.025moles` of
`NH_3`

Mass of
`LiBr=moles* {\text {Molar mass}}=0.025moles* 86.8g/mol=2.17g`

Thus 2.17 g of bromide product would be formed