Question

5) Determine the mass if lithium hydroxide that is produced when 12.87 g of lithium

Answer 1

Answer: 26.54 grams

Explanation:

To calculate the moles :

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

$\text{Moles of lithium nitride}=\frac{12.87g}{34.83g/mol}=0.369moles$

$Li_3N+3H_2O\rightarrow 3LiOH+NH_3$  

$Li_3N$ is the limiting reagent as it limits the formation of product and $H_2O$ is the excess reagent

According to stoichiometry :

As 1 moles of $Li_3N$ give = 3 moles of $LiOH$

Thus 0.369 moles of $O_2$ give =$\frac{3}{1}\times 0.369=1.108moles$  of $LiOH$

Mass of $LiOH=moles\times {\text {Molar mass}}=1.108moles\times 23.95g/mol=26.54g$

Thus  26.54 g of $LiOH$ will be produced from the given mass.