How many grams of Mg(OH)2 can be prepared from the reaction of 20.0g of Mg3N2 with 10.0g of H2O?

To answer this question, we need to determine the limiting reactant in the given reaction and then calculate the amount of Mg(OH)2 that can be produced from the limiting reactant.

The balanced chemical equation for the reaction is as follows:
Mg3N2 + 6H2O -> 3Mg(OH)2 + 2NH3

First, let's calculate the number of moles of each reactant using the molar mass:

Molar mass of Mg3N2 = 3x(24.3 g/mol of Mg) + 2x(14.0 g/mol of N) = 100.9 g/mol
Number of moles of Mg3N2 = 20.0 g / 100.9 g/mol = 0.198 moles

Molar mass of H2O = 2x(1.0 g/mol of H) + 16.0 g/mol of O = 18.0 g/mol
Number of moles of H2O = 10.0 g / 18.0 g/mol = 0.556 moles

Next, let's compare the stoichiometric ratio of Mg3N2 to H2O in the balanced equation:
1 mole of Mg3N2 reacts with 6 moles of H2O

From the above ratio, we can see that 1 mole of Mg3N2 requires 6 moles of H2O. Therefore, to react fully, 0.198 moles of Mg3N2 would require 6 times that amount of H2O:
0.198 moles of Mg3N2 x 6 moles of H2O/mole of Mg3N2 = 1.188 moles of H2O

Since we have only 0.556 moles of H2O, which is less than 1.188 moles required, H2O is the limiting reactant. This means that all the H2O will be used up, and Mg3N2 will be in excess.

Now, let's calculate the mass of Mg(OH)2 formed from the limiting reactant:

Molar mass of Mg(OH)2 = 24.3 g/mol of Mg + 2x(1.0 g/mol of H) + 16.0 g/mol of O = 58.3 g/mol
Number of moles of Mg(OH)2 formed = 0.556 moles of H2O x 3 moles of Mg(OH)2/6 moles of H2O = 0.278 moles

Mass of Mg(OH)2 formed = 0.278 moles x 58.3 g/mol = 16.2 g

Therefore, from the reaction of 20.0 g of Mg3N2 with 10.0 g of H2O, you can prepare 16.2 grams of Mg(OH)2.