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 (NaN_{3}) into its component elements inflates automotive air bags.

Given the following chemical decomposition of sodium azide:

2NaN_{3}(s) ----> 2Na(s) + 3N_{2}(g).

(a) How many moles of N_{2} are produced by the decomposition of 1.500 moles of NaN_{3}?

(b) Calculate grams of NaN_{3} needed to produce 10 g of N_{2}.

(c) Calculate grams of NaN_{3} needed to produce 10 ft^{3} of nitrogen gas if the gas has a density of 1.25 g/L.

**The strings**:
S_{7}P_{6}A_{66} (interaction - chemical).

S_{7}P_{1}A_{15} (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 NaN_{3} reacts with 3 moles of N_{2}

So, 1.500 moles of NaN_{3} will react with (1.500 x 3)/2 = 2.25 moles of N_{2}
**(b)**Molar mass of nitrogen = 14 x 2 = 28 g

28 g is weight of 1 mole of N_{2}

So, 10 g is the weight of 10/28 moles = 0.357 moles of N_{2}

3 moles of N_{2} is formed by 2 moles of NaN_{3}

So, 0.357 mol of N_{2} is formed by (0.357 x 2)/3 = 0.238 mol of moles of NaN_{3}

Molar mass of moles of NaN_{3} = 23 + 42 = 65 g

1 mole of NaN_{3} weighs 65 g

So, 0.238 of NaN_{3} weighs (0.238 x 65) = 15.5 g.
**(c)** 1 Liter = 0.0353 ft^{3}

So, 10 ft^{3} of N_{2} = 10/0.0353 Liters = 283 L.

Mass of 283 L of N_{2} = 1.25 x 283 = 353.75 g

Molar mass of N_{2} = 14 x 2 = 28

28 g of N_{2} is the weight of 1 mole of N_{2}

So, 353.75 g is the weight of 353.75/28 = 12.64 moles

3 moles of N_{2} is formed from 2 moles of NaN_{3}

So, 12.64 moles of N_{2} is formed by (2 x 12.64)/3 oles of NaN_{3} = 8.43 moles

Molar mass of NaN_{3} = 23 + 42 = 65

1 mole of NaN_{3} weighs 65 g

So, 8.43 moles of NaN_{3} weighs 8.43 x 65 = 548 g.

Math

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