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Sagot :
Answer: 2.17 g of bromide product would be formed
Explanation:
The reaction of calcium bromide with lithium oxide will be:
[tex]CaBr_2+Li_2O\rightarrow 2LiBr+CaO[/tex]
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of calcium bromide}=\frac{2.5g}{199.9g/mol}=0.0125moles[/tex]
As lithium oxide is in excess, calcium bromide is the limiting reagent.
According to stoichiometry :
1 mole of [tex]CaBr_2[/tex] produce = 2 moles of [tex]LiBr[/tex]
Thus 0.0125 moles of [tex]CaBr_2[/tex] will require=[tex]\frac{2}{1}\times 0.0125=0.025moles[/tex] of [tex]NH_3[/tex]
Mass of [tex]LiBr=moles\times {\text {Molar mass}}=0.025moles\times 86.8g/mol=2.17g[/tex]
Thus 2.17 g of bromide product would be formed
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