A 3.82 g sample of magnesium nitride is reacted with 7.73 g of water. 푀푔 3 푁 2 + 3퐻 2 푂→2푁퐻 3 + 3푀푔푂 The actual yield of MgO is 3.60 g. What is the percentage yield of the reaction?

To find the percentage yield of the reaction, we first need to find the theoretical yield.

Step 1: Find the molar mass of the reactants and products.

• Molar mass of Mg3N2 = 3(24.31 g/mol Mg) + 2(14.01 g/mol N) = 100.95 g/mol
• Molar mass of H2O = 2(1.01 g/mol H) + 16.00 g/mol O = 18.02 g/mol
• Molar mass of MgO = 24.31 g/mol Mg + 16.00 g/mol O = 40.31 g/mol

Step 2: Convert the mass of the reactants to moles.

• Moles of Mg3N2 = 3.82 g / 100.95 g/mol = 0.0378 mol
• Moles of H2O = 7.73 g / 18.02 g/mol = 0.429 mol

Step 3: Determine the limiting reactant. From the balanced equation, we know that 1 mol of Mg3N2 reacts with 3 mol of H2O to produce 3 mol of MgO. Therefore, the amount of MgO that can be produced is limited by the amount of Mg3N2 and H2O we have.

• If all the Mg3N2 reacts, it will produce 0.0378 mol * 3 = 0.1134 mol of MgO.
• If all the H2O reacts, it will produce 0.429 mol / 3 = 0.143 mol of MgO.

Since Mg3N2 produces less MgO, it is the limiting reactant.

Step 4: Calculate the theoretical yield of MgO. Theoretical yield = 0.1134 mol * 40.31 g/mol = 4.57 g

Step 5: Calculate the percentage yield. Percentage yield = (actual yield / theoretical yield) * 100% Percentage yield = (3.60 g / 4.57 g) * 100% = 78.77%

So, the percentage yield of the reaction is approximately 78.77%.