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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.
S7P6A66 (interaction - chemical).
S7P1A15 (containership - mass).
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.
The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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