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mg3n2 + 3k2o --> 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?

There will be 525.2 grams of K3N produced

Explanation:

Step 1: Data given

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

Magnesium nitride (Mg3N) = in excess

Molar mass of K3N = 131.3 g/mol

Step 2: The balanced equation

Mg3N2 + 3K2O → 3MgO + 2K3N

Step 3: Calculate moles of K3N

The limiting reactant is K2O.

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

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

Step 4: Calculate mass of K3N

Mass of K3N = moles K3N * molar mass K3N

Mass of K3N = 4 moles * 131.3 g/mol

Mass of K3N = 525.2 grams

There will be 525.2 grams of K3N produced

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the product of the pressure and volume is also constant