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

Explanation:

The reaction of calcium bromide with lithium oxide will be:

$CaBr_2+Li_2O\rightarrow 2LiBr+CaO$

To calculate the moles :

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

$\text{Moles of calcium bromide}=\frac{2.5g}{199.9g/mol}=0.0125moles$

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=$\frac{2}{1}\times 0.0125=0.025moles$  of $NH_3$

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

Thus 2.17 g of  bromide product would be formed