A3.82-g sample of magnesium nitride is reacted with 7.73 g of water. mg3n2 (s) + 3 h2o (â) â 2 nh3 (g) + 3 mgo (s) the yield of mgo is 3.60 g. what is the percent yield in the reaction?

The reaction is properly written as

Mg₃N₂ (s) + 3 H₂O (l) --> 2 NH₃ (g) + 3 MgO (s) 

Molar mass of Mg₃N₂ = 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 = 78.77%

The correct answer is c. orbital shape. "the angular momentum quantum number l describes the shape of an orbital. it corresponds to each orbital type, i.e. (0,1,2,−1)=(s,p,d,f,g,"

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