Question

Consider the reaction of Mg3N2 with H2O to form Mg(OH)2 and NH3. If 3.41 g Mg3N2 is reacted with excess H2O and 4.11 g of Mg(OH)2 is ultimately isolated, what is the percent yield for the reaction?

Answer 1

The percent yield for the reaction is 89.4%.

Step-by-step explanation:

To calculate the percent yield for the reaction, we need to compare the actual yield (the amount of product obtained in the experiment) to the theoretical yield (the amount of product that should have been obtained according to the balanced equation).

First, we calculate the theoretical yield of Mg(OH)2 using stoichiometry. From the balanced equation, we know that 1 mol of Mg3N2 reacts with 6 mol of H2O to produce 3 mol of Mg(OH)2. So, the molar mass of Mg3N2 is 100.95 g/mol, and the molar mass of Mg(OH)2 is 58.33 g/mol.

Next, we calculate the actual yield of Mg(OH)2 by dividing the mass of Mg(OH)2 obtained by the molar mass of Mg(OH)2 and multiplying by 100 to get the percent yield.

Percent yield = (actual yield / theoretical yield) * 100

Percent yield = (4.11 g / (100.95 g/mol * (3.41 g / 100.95 g/mol)) * 58.33 g/mol) * 100

Percent yield = 89.4%