Question

Sodium azide (NaN3) yields N2 gas when heated to 300 ∘C, a reaction used in automobile air bags.

If 1.00 mol of N2 has a volume of 47.0 L under the reaction conditions, how many liters of gas can be formed by heating 39.0 g of NaN3? The reaction is:
2NaN3→3N2(g)+2Na

Answer 1

Answer on the question #54713 – Chemistry – General chemistry

Question:

Sodium azide (NaN₃) yields N₂ gas when heated to 300 °C, a reaction used in automobile air bags.

If 1.00 mol of N₂ has a volume of 47.0 L under the reaction conditions, how many liters of gas can be formed by heating 39.0 g of NaN₃? The reaction is:

$$2 \mathrm{NaN_3} \rightarrow 3 \mathrm{N_2(g)} + 2 \mathrm{Na}$$

Answer:

According to the reaction equation, the number of the moles of the sodium azide and nitrogen relates as:

$$\frac{n(\mathrm{NaN_3})}{2} = \frac{n(\mathrm{N_2})}{3}$$

The number of the moles of the sodium azide is:

$$n(\mathrm{NaN_3}) = \frac{m(\mathrm{NaN_3})}{M(\mathrm{NaN_3})} = \frac{39.0}{65} = 0.6 \text{ mol}$$

Then, the number of the moles of nitrogen gas is:

$$n(\mathrm{N_2}) = n(\mathrm{NaN_3}) * \frac{3}{2} = 0.6 * \frac{3}{2} = 0.9 \text{ mol}$$

The volume of nitrogen is:

$$V(\mathrm{N_2}) = n(\mathrm{N_2}) * 47 = 42.3 \text{ L}$$

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