If 29.0g of potassium reacts with 66.5g of barium nitride, what mass of potassium nitride will be produced?

32.4 grams of K3N will be produced

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