Answer
V= 1.8136 L
Procedure
The equation is already balanced
Mg₃N₂ + 6H₂O -> 3Mg(OH)₂ + 2NH₃
1. Use stoichiometry to get the moles of ammonia
[tex]6.0\text{ g Mg}_3\text{N}_2\text{ }\frac{1\text{ mol Mg}_3\text{N}_2}{148.3\text{ g Mg}_3\text{N}_2}\frac{2\text{ mol NH}_3}{1\text{ mol Mg}_3\text{N}_2}=0.081\text{ mol NH}_3[/tex]2. Use the ideal gas equation considering STP conditions to get the volume
[tex]PV=nRT[/tex]
Solving for V
[tex]V=\frac{nRT}{P}=\frac{0.081\text{ mol }0.082(\text{ L. atm. mol}^{-1}\text{. }\degree\text{K}^{-1})\text{ }273.15\degree K}{1\text{ atm}}=1.8136\text{ L}[/tex]
