2. What is the percent yield of Mg3N2 if 73.9 g of Mg reacts with 31.6 g of N2 to produce 79 g of Mg3N2 according to the following chemical reaction? (Reaction may or may not be balanced.)
mg+n2=Mg3N2
This is a limiting reagent problem. You know that because amounts are given for BOTH reactants.
3Mg + N2 ==> Mg3N2
Convert 73.9g Mg to mols. mol = grams/atomic mass.
Do the same for 31.6 g N2.
Using the coefficients in the balanced equation, convert mols Mg to mols Mg3N2.
Do the same for mols N2 to mols Mg3N2.
It is likely that the two values for mols Mg3N2 will not agree which means one of them is not right; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Use the smaller value and convert mol to grams. g = mols x molar mass. This is the theoretical yield (TY).
%yield = (actual yield/TY)*100 = ?
To calculate the percent yield of Mg3N2, you'll need to compare the actual amount of Mg3N2 obtained (79g) with the theoretical amount that should have been obtained if the reaction went to completion.
Step 1: Write and balance the chemical equation:
The given equation is already in unbalanced form. To balance it, let's ensure equal atoms on both sides:
3 Mg + N2 → Mg3N2
Step 2: Calculate the molar masses:
Mg: 24.31 g/mol
N2: 28.01 g/mol
Mg3N2: 100.93 g/mol
Step 3: Determine the limiting reactant:
To determine the limiting reactant, we'll need to calculate the number of moles of each reactant using their molar masses.
Moles of Mg = mass (g) / molar mass (g/mol)
Moles of Mg = 73.9 g / 24.31 g/mol = 3.04 mol
Moles of N2 = mass (g) / molar mass (g/mol)
Moles of N2 = 31.6 g / 28.01 g/mol = 1.13 mol
Based on the balanced equation, we can see that the mole ratio between Mg and Mg3N2 is 3:1, while the mole ratio between N2 and Mg3N2 is 1:1. Therefore, the limiting reactant is N2 because it produces fewer moles of Mg3N2.
Step 4: Calculate the theoretical yield of Mg3N2:
Using the stoichiometry of the balanced equation, we know that 1 mole of N2 reacts to produce 1 mole of Mg3N2.
Moles of Mg3N2 = Moles of N2 = 1.13 mol
Mass of Mg3N2 = Moles of Mg3N2 × molar mass of Mg3N2
Mass of Mg3N2 = 1.13 mol × 100.93 g/mol = 114.42 g
Step 5: Calculate the percent yield:
Percent yield = (Actual yield / Theoretical yield) × 100%
Percent yield = (79 g / 114.42 g) × 100% ≈ 69.05%
The percent yield of Mg3N2 in this reaction is approximately 69.05%.