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%

yes, water is wet, just as it makes other things wet, it is wet itself, kinda like ice, when ice is in water, the ice makes water cold, and ice is cold, therefore with that example, water is wet