76.2k views
3 votes
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

1 Answer

5 votes

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.

User Mark Van Straten
by
8.9k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.