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aleksandrvk [35]
3 years ago
14

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

Chemistry
1 answer:
Ede4ka [16]3 years ago
3 0

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

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