The percent yield of this reaction is calculated as follows
Mg3N2 + 3H2O =2NH3 + 3Mgo
calculate the theoretical yield,
moles=mass/molar mass
moles Mg3N2= 3.82 g/100g/mol= 0.0382 moles(limiting regent)
moles of H2o= 7.73g/18g/mol = 0.429 moles ( in excess_)
by use of mole ratio between Mg3N2 to MgO which is 1:3 the moles of MgO = 0.0382 x3 = 0.1146 moles
mass =moles x molar mass
the theoretical mass is therefore = 0.1146mole x 40 g/mol = 4.58 grams
The % yield = actual mass/theoretical mass x1000
= 3.60/4.584 x100= 78.5%
A) Calcium Dihydrogen Phosphate
B) Iron(II) Sulfate
C) Calcium Carbonate
D) Magnesium Oxide
E) Sodium Nitrite
F) Potassium iodide
