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?

Given

Mass of magnesium nitride: 3.8 g

Mass of water: 3.3 g

Mass of mgO : 3.6 g

Mg3N2 + 3 H2O --> 2 NH3+ 3 MgO

Now we know that moles=

mass of the sample/ molecular mass of the sample

Moles of Mg3N2=

given mass of Mg3N2/molecular mass of Mg3N2

Moles of Mg3N2 = 3.8/101=0.038

Moles of H2O=

given mass of H2O /molecular mass of H2O

Moles of H2O = 3.3 /18= 0.18

According to the equation since there are 3 entities of H2O we get number of moles equals 0.55.

Moles of MgO from the equation we can calculate as

Moles of MgO = 0.038 x3=0.114

Thus the mass of MgO can be calculated as no of molesx molecular mass

Mass of MgO= 0.114 x 40.33=4.59g (This is the theoretical amount present).

Actual amount present equals 3.6 g

Thus the% yield = (actual amount)/(theoretical amount) x 100

% yield= (3.6/4.59) x 100= 78.42

the answer is 1 i believe