Question
Sodium azide (NaN $_{3} )$ yields $\mathrm{N}_{2}$ gas when heated to $300^{\circ} \mathrm{C},$ a reaction used in automobile air bags. If 1.00 $\mathrm{mol}$ of $\mathrm{N}_{2}$ has a volume of 47.0 $\mathrm{L}$ under the reaction conditions, how many liters of gas can be formed by heating 38.5 $\mathrm{g}$ of $\mathrm{NaN}_{3} ?$ The reaction is$$2 \mathrm{NaN}_{3} \longrightarrow 3 \mathrm{N}_{2}(g)+2 \mathrm{Na}$$
Step 1
We do this by dividing the given mass by the molar mass of Sodium azide. The molar mass of Sodium azide is 65.1 g/mol. So, we have: $$\frac{38.5 \, \text{g}}{65.1 \, \text{g/mol}} = 0.591 \, \text{mol} \, \text{of NaN}_{3}$$ Show more…
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Sodium azide $\left(\mathrm{NaN}_{3}\right)$ yields $\mathrm{N}_{2}$ gas when heated to $300{ }^{\circ} \mathrm{C}$, a reaction used in automobile air bags. If $1.00 \mathrm{~mol}$ of $\mathrm{N}_{2}$ has a volume of $47.0 \mathrm{~L}$ under the reaction conditions, how many liters of gas can be formed by heating $38.5 \mathrm{~g}$ of $\mathrm{NaN}_{3} ?$ The reaction is $$ 2 \mathrm{NaN}_{3}(s) \longrightarrow 3 \mathrm{~N}_{2}(g)+2 \mathrm{Na}(s) $$
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During a collision, automobile air bags are inflated by the $\mathrm{N}_{2}$ gas formed by the explosive decomposition of sodium azide, $\mathrm{NaN}_{3}$ $$2 \mathrm{NaN}_{3} \longrightarrow 2 \mathrm{Na}+3 \mathrm{~N}_{2}$$ What mass of sodium azide would be needed to inflate a 30.0-L bag to a pressure of $1.40 \mathrm{~atm}$ at $25^{\circ} \mathrm{C}$ ?
Sodium azide, the explosive compound in automobile air bags, decomposes according to the following equation: $$ 2 \mathrm{NaN}_{3}(\mathrm{s}) \longrightarrow 2 \mathrm{Na}(\mathrm{s})+3 \mathrm{N}_{2}(\mathrm{g}) $$ What mass of sodium azide is required to provide the nitrogen needed to inflate a $75.0-\mathrm{L}$. bag to a pressure of 1.3 atm at $25^{\circ} \mathrm{C} ?$
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