Question

In the following reaction 2NaN3 + 2Na + 3N2 , if 750 g of NaN3 decompose to form
265.20 g of Na, how much N2 is produced?
(a) 484.80 g
(b) 576.80 g
(c) 357.80 g
(d) 464.80 g​

Answer 1

Answer:

357.80 this answers okkkk

Answer 2

242.307 g N₂ is produced.

The given reaction is,

2 NaN₃ ⇄ 2 Na + 3 N₂

The molecular weight of NaN₃ is = {23 + (3 × 14)}

                                                       = 23 + 42

                                                       = 65

The molecular weight of Na is = 23

The molecular weight of N₂ i s= 14

Given,

750 g NaN₃ is decomposed.

From the equation, we get,

(2 × 65) g NaN₃ gives (3 × 14) g of N₂

∴ 1 g NaN₃ gives (3 × 14)/ (2 × 65) g of N₂

∴ 750 g NaN₃ gives (3 × 14 × 750)/ (2 × 65) g of N₂

                              = 242.307 g of N₂

So, 242.307 g of N₂ is produced.