Automotive air bags inflate when sodium azide, \(\mathrm{NaN}_{3}\) , rapidly decomposes to its component elements: $$ 2 \mathrm{NaN}_{3}(s) \longrightarrow 2 \mathrm{Na}(s)+3 \mathrm{N}_{2}(g) $$ (a) How many moles of \(\mathrm{N}_{2}\) are produced by the decomposition of 1.50 \(\mathrm{mol}\) of \(\mathrm{NaN}_{3} ?\) (b) How many grams of NaN \(_{3}\) are required to form 10.0 \(\mathrm{g}\) of nitrogen gas? (c) How many grams of NaN \(_{3}\) are required to produce 10.0 \(\mathrm{ft}^{3}\) of nitrogen gas, about the size of an automotive air bag, if the gas has a density of 1.25 \(\mathrm{g} / \mathrm{L} ?\)

From the balanced chemical equation, $$ 2 \mathrm{NaN}_{3}(s) \longrightarrow 2 \mathrm{Na}(s)+3 \mathrm{N}_{2}(g), $$ we can see that every 2 moles of NaN3 will decompose to produce 3 moles of N2 gas. This means that the mole ratio between N2 and NaN3 can be expressed as: $$ \frac{3 \ \text{moles of} \ \mathrm{N}_{2}}{2 \ \text{moles of} \ \mathrm{NaN}_{3}}. $$

Using the mole ratio from Step 1, we can determine the moles of N2 produced by 1.50 moles of NaN3: $$ \text{Moles of N2} = 1.50 \ \text{moles of} \ \mathrm{NaN}_{3} \cdot \frac{3 \ \text{moles of} \ \mathrm{N}_{2}}{2 \ \text{moles of} \ \mathrm{NaN}_{3}} = \boxed{2.25 \ \text{moles of} \ \mathrm{N}_{2}}. $$

To answer questions (b) and (c), we need the molar mass of the NaN3 and N2 gases. Molar mass of NaN3: $$ \text{NaN}_{3} = 1 \times (\text{Na}) + 3 \times (\text{N}). $$ From the periodic table, we know the atomic masses of Sodium (Na) and Nitrogen (N) are approximately 23 g/mol and 14 g/mol, respectively. Thus, the molar mass of NaN3 is: $$ \text{Molar Mass of NaN}_{3} = 1 \times (23 \ \text{g/mol}) + 3 \times (14 \ \text{g/mol}) = 23 + 42 = 65 \ \text{g/mol}. $$ Molar mass of N2: $$ \text{N}_{2} = 2 \times (\text{N}). $$ Knowing the atomic mass of Nitrogen (N) is approximately 14 g/mol: $$ \text{Molar Mass of N}_{2} = 2 \times (14 \ \text{g/mol}) = 28 \ \text{g/mol}. $$

Using the mole ratio from Step 1 and the molar masses calculated in Step 3, we can determine the grams of NaN3 needed to form 10.0 g of N2: $$ \text{Grams of NaN}_{3} = \frac{10.0 \ \text{g of} \ \mathrm{N}_{2}}{28 \ \text{g/mol}} \cdot \frac{3 \ \text{moles of} \ \mathrm{N}_{2}}{2 \ \text{moles of} \ \mathrm{NaN}_{3}} \cdot 65 \ \text{g/mol} \ = \boxed{34.4 \ \text{g of} \ \mathrm{NaN}_{3}}. $$

First, let's convert 10 ft³ of N2 gas into liters. We know that 1 ft³ = 28.317 L: $$ \text{Volume of N}_{2}\text{ in liters} = 10.0 \ \text{ft}^3 \cdot \frac{28.317 \ \text{L}}{1 \ \text{ft}^3} = 283.17 \ \text{L}. $$ Next, we can determine the mass of N2 gas in grams using its density (1.25 g/L): $$ \text{Mass of N}_{2} = 283.17 \ \text{L} \cdot 1.25 \ \text{g/L} = 354 \ \text{g}. $$ Finally, using the mole ratio from Step 1 and the molar masses calculated in Step 3, we can determine the grams of NaN3 needed to produce 354 g of N2: $$ \text{Grams of NaN}_{3} = \frac{354 \ \text{g of} \ \mathrm{N}_{2}}{28 \ \text{g/mol}} \cdot \frac{3 \ \text{moles of} \ \mathrm{N}_{2}}{2 \ \text{moles of} \ \mathrm{NaN}_{3}} \cdot 65 \ \text{g/mol} \ = \boxed{1236 \ \text{g of} \ \mathrm{NaN}_{3}}. $$