When a 3.80 g sample of magnesium nitride (MW 101g/mol) is reacted with 3.30 g of water, 3.60 g of MgO is formed. What is the percent yield of this reaction? Mg3N2 + 3 H2O - - > 2 NH3 + 3 MgO

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%