How many liters of nitrogen gas are needed to react with 90.0 g of potassium at STP in order to produce potassium nitride according to the following reactions?

The atomic mass of potassium is 19 g/mol

90.0 g of potassium contains;

= 90.0/19

= 4.737 moles

The reaction between potassium and nitrogen is given by the equation;

6K + N2 = 2K3N

The mole ratio of potassium and nitrogen is 6:1

The number of moles of nitrogen;

= 4.737/6

= 0.7895 moles

But, 1 mole of nitrogen occupies 22.4 liters at STP

Therefore; 0.7895 moles * 22.4 liters

= 17.685 liters