Mg3N2 + 3K2O à 3MgO + 2K3N If 6 moles of potassium oxide reacts with an excess amount of magnesium nitride, how many grams of potassium nitride will be made? Question 3 options: 524g K3N 1524 g K3N 14 moles K3N 345g MgO

There is 524 grams of K3N produced

Explanation:

Step 1: Data given

Number of moles of potassium oxide (K2O) = 6 moles

Magnesium nitride (Mg3N2) = in excess

Step 2: The balanced equation

Mg3N2 + 3K2O → 3MgO + 2K3N

Step 3: Calculate moles of K3N

The limiting reactant is potassium oxide

For 1 mol of Mg3N2 we need 3 moles of K2O to produce 3 moles of MgO and 2 moles of K3N

For 6 moles of K2O we'll have 2/3 * 6 = 4 moles K3N

Step 4: Calculate mass of K3N

Mass K3N = 4 moles * 131 g/mol

Mass K3N = 524 grams

There is 524 grams of K3N produced

524g K₃N

Explanation:

The balanced reaction:

Mg₃N₂ + 3K₂O → 3MgO + 2K₃N

If the information told us, that the excess is in the Mg₃N₂, the K₂O is the limiting reagent.

3 moles of K₂O produce 2 moles of K₃N

6 moles of K₂O will produce (6. 2) / 3 = 4 moles

Molar weight of K₃N = 131.3 g/m

4 m. 131.3 g/m = 525.2 g