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%

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