Calculations Of Moles And Grams When Sodium Azide Decomposes To Sodium And Nitrogen
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Calculations Of Moles And Grams When Sodium Azide Decomposes To Sodium And Nitrogen The rapid decomposition of sodium azide (NaN3) into its component elements inflates automotive air bags.
Given the following chemical decomposition of sodium azide:

2NaN3(s) ----> 2Na(s) + 3N2(g).

(a) How many moles of N2 are produced by the decomposition of 1.500 moles of NaN3?
(b) Calculate grams of NaN3 needed to produce 10 g of N2.
(c) Calculate grams of NaN3 needed to produce 10 ft3 of nitrogen gas if the gas has a density of 1.25 g/L.

The strings: S7P6A66 (interaction - chemical).
S7P1A15 (containership - mass).

The math:
Although we are calculating , mass, grams and moles, the chemical reaction is the reason for the calculation. So Pj Problem of Interest is of type interaction (chemical). (a) 2 moles of NaN3 reacts with 3 moles of N2
So, 1.500 moles of NaN3 will react with (1.500 x 3)/2 = 2.25 moles of N2

(b)Molar mass of nitrogen = 14 x 2 = 28 g
28 g is weight of 1 mole of N2
So, 10 g is the weight of 10/28 moles = 0.357 moles of N2
3 moles of N2 is formed by 2 moles of NaN3
So, 0.357 mol of N2 is formed by (0.357 x 2)/3 = 0.238 mol of moles of NaN3
Molar mass of moles of NaN3 = 23 + 42 = 65 g
1 mole of NaN3 weighs 65 g
So, 0.238 of NaN3 weighs (0.238 x 65) = 15.5 g.

(c) 1 Liter = 0.0353 ft3
So, 10 ft3 of N2 = 10/0.0353 Liters = 283 L.
Mass of 283 L of N2 = 1.25 x 283 = 353.75 g
Molar mass of N2 = 14 x 2 = 28
28 g of N2 is the weight of 1 mole of N2
So, 353.75 g is the weight of 353.75/28 = 12.64 moles
3 moles of N2 is formed from 2 moles of NaN3
So, 12.64 moles of N2 is formed by (2 x 12.64)/3 oles of NaN3 = 8.43 moles
Molar mass of NaN3 = 23 + 42 = 65
1 mole of NaN3 weighs 65 g
So, 8.43 moles of NaN3 weighs 8.43 x 65 = 548 g.

Math

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