Answer:
32.4 grams of K3N will be produced
Step-by-step explanation:
Step 1: Data given
Mass of potassium = 29.0 grams
Mass of barium nitride = 66.5 grams
Atomic mass of potassium = 39.10 g/mol
Molar mass of barium nitride = 440 g/mol
Step 2: The balanced equation
6K + Ba3N2 → 2K3N + 3Ba
Step 3: Calculate moles
Moles = mass / molar mass
Moles K = 29.0 grams / 39.10 g/mol
Moles K = 0.742 moles
Moles Ba3N2 = 66.5 grams / 440 g/mol
Moles Ba3N2 = 0.151 moles
Step 4: Calculate the limiting reactant
For 6 moles K we need 1 mol Ba3N2 to produce 2 moles K3N and 3 moles Ba
K is the limiting reactant. It will completely be consumed (0.742 moles). Ba3N2 is in excess. There will react 0.742/ 6 = 0.124 moles
There will remain 0.151 - 0.124 = 0.027 moles
Step 5: Calculate moles K3N
For 6 moles K we need 1 mol Ba3N2 to produce 2 moles K3N and 3 moles Ba
For 0.742 moles K we'll have 0.742/3 = 0.247 moles K3N
Step 6: Calculate mass K3N
Mass K3N = moles K3N * molar mass K3N
Mass K3N = 0.247 moles * 131.3 g/mol
Mass K3N = 32.4 grams
32.4 grams of K3N will be produced