What mass of water is needed to react with 32.9 g Li3N?

Answers

 moles Li3N = 98.7 g/34.8297 g/mol=2.83 

the ratio between Li3N and H2O is 1 : 3 
moles water neeeded = 2.83 x 3=8.49 
mass water = 8.49 mol x 18.02 g/mol=153.0 g 

moles NH3 produced = 2.83 
molescules of NH3 = 2.83 x 6.02 x 10^23=1.70 x 10^24 

moles NH3 (STP) = 45.0/22.4 =2.0 
moles Li3N = 2.0 
mass Li3N = 2.0 x 34.8297 g/mol=69.6 g

Mass = 51.12 g

Explanation:

Given data:

Mass of lithium nitride = 32.9 g

Mass of water needed = ?

Solution:

Chemical equation:

Li₃N + 3H₂O   → NH₃ + 3LiOH

Number of moles of lithium nitride:

Number of moles = mass/ molar mass

Number of moles = 32.9 g/ 34.83 g/mol

Number of moles = 0.945 mol

Now we will compare the moles of water and lithium nitride.

                         Li₃N               :                H₂O

                           1                   :                  3

                           0.945          :                 3×0.945 = 2.84 mol

Mass of water:

Mass = number of moles × molar mass

Mass = 2.84 mol × 18 g/mol

Mass = 51.12 g

The answer to your question is 50.76 g of H₂O

Explanation:

Data

mass of water = ?

mass of Li₃N = 32.9 g

Process

1.- Write the balanced chemical reaction

                    Li₃N  +  3H₂O   ⇒  3LiOH   +  NH₃

2.- Calculate the molar mass of Li₃N and H₂O

Li₃N = (3 x 7) + (1 x 14) = 35 g

H₂O = (2 x 1) + (16 x 1) = 18 g

3.- Use proportions to calculate the mass of water

                   35 g of Li₃N (3 x 18) g of water

                    32.9 g of Li₃N x

                    x = (32.9 x 54)/35

                    x = 1776.6/35

                    x = 50.76 g of H₂O



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