Consider the reaction of Mg3N2 with H2O to form Mg (OH) 2 and NH3. If 4.33 g H2O is reacted with excess Mg3N2 and 6.26 g of Mg (OH) 2 is ultimately isolated, what is the percent yield for the reaction?

89.34%

Explanation:

First, write a balanced reaction.

Mg3N2 + 6H2O - -> 3Mg (OH) 2 + 2NH3

Next determine the moles of the known substance, or limiting reagent (H2O)

n = m/MM

n (H2O) = 4.33 / (1.008*2) + 16

n (H2O) = 0.2403

Use the mole ratio to find the moles of Mg (OH) 2

0.2403 : 2

n (Mg (OH) 2) = 0.1202

Next, find the theoretical mass of Mg (OH) 2 that should have been produced

m = n * MM

m = 0.1202 * (24.305 + (16*2) + (1.008 * 2))

=7.007g

To find percentage yield, divide the experimental amount by the theoretical amount and multiply by 100.

6.26 / 7.007 * 100

=89.34%