Question

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

nitride reacts with an excess of water according to the following process.
Li:N + 3 H2O → 3 LiOH + NH3

Answer 1

Answer: 26.54 grams

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of lithium nitride}=(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 =
`(3)/(1)* 0.369=1.108moles` of
`LiOH`

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

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