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