The reaction is properly written as
Mg₃N₂ (s) + 3 H₂O (l) --> 2 NH₃<span> (g) + 3 MgO (s)
Molar mass of Mg</span>₃N₂ = 100.95 g/mol
Molar mass of H₂O = 18 g/mol
Molar mass of MgO = 40.3 g/mol
Moles Mg₃N₂: 3.82/100.95 = 0.0378
Moles H₂O: 7.73/18 = 0.429
Theo H₂O required for available Mg₃N₂: 0.0378*3/1 = 0.1134 mol
Hence, the limiting reactant is Mg₃N₂.
Thus,
Theoretical Yield = 0.0378 mol Mg₃N₂ * 3 mol MgO/Mg₃N₂ * 40.3 g/mol
Theo Yield = 4.57 g
Percent Yield = Actual Yield/Theo Yield * 100
Percent Yield = 3.60 g/4.57 g * 100 =<em> 78.77%</em>
Answer:
.371 mole of NaCl
Explanation:
Na Cl Mole weight = 22.989 + 35.45 = 58.439 g/mole
21.7 g / 58.439 g/mole = .371 mole
Red is the answer in the blank
1.) homogeneous mixtures
2.) <span>NaCI, because sodium is a metal and chlorine is a nonmetal</span>
3.) I'm thinking the first the option
Average atomic mass listed for nitrogen in the periodic table is 14
Hope this helps!
