a)
Step 1
The reaction provided:
Li3N + 3H20 => NH3 + 3LiOH (completed and balanced)
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Step 2
Information provided:
40 g of H2O
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Information needed:
The molar mass of Li3N = 34.9 g/mol approx.
The molar mass of H2O = 18.0 g/mol approx.
(please, use your periodic table)
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Step 3
By stoichiometry,
Procedure:
Li3N + 3H20 => NH3 + 3LiOH
34.9 g L3N --------------- 3 x 18.0 g H2O
X --------------- 40 g H2O
X = (40 g H2O x 34.9 g L3N)/(3 x 18 g H2O)
X = 25.9 g
Answer: 25.9 g of L3N will react with H2O