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

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