How many grams of sodium metal is produced when 744 ml of nitrogen gas at stp is also produced?

             0.50 g of Na Metal

Solution:
The equation for given reaction is as follow,

                                   2 NaN₃    →     2 Na  +   3 N₂

According to equation,
When,

    67.2 L (3 mole) N₂ is produced along with  =  46 g (2 mole) of Na Metal
So,
            0.744 L N₂ produced will accompany  =  X g of Na Metal

Solving for X,
                      X  =  (0.744 L × 46 g) ÷ 67.2 L

                      X  =  0.50 g of Na Metal

The answer is b metal denting

The answer to this is the 3rd choice!