To determine how many grams of [tex]\( \text{NaN}_3 \)[/tex] (sodium azide) are required to produce 10.0 liters of nitrogen gas (N[tex]\(_2\)[/tex]) with a density of 1.25 g/L, follow these steps:
1. Calculate the mass of nitrogen gas:
[tex]\[
\text{mass of nitrogen gas} = \text{volume} \times \text{density}
\][/tex]
Given:
[tex]\[
\text{volume of nitrogen gas} = 10.0 \, \text{L}
\][/tex]
[tex]\[
\text{density of nitrogen gas} = 1.25 \, \text{g/L}
\][/tex]
[tex]\[
\text{mass of nitrogen gas} = 10.0 \, \text{L} \times 1.25 \, \text{g/L} = 12.5 \, \text{g}
\][/tex]
2. Convert the mass of nitrogen gas to moles:
The molar mass of nitrogen gas (N[tex]\(_2\)[/tex]) is 28 g/mol.
[tex]\[
\text{moles of nitrogen gas} = \frac{\text{mass}}{\text{molar mass}}
\][/tex]
[tex]\[
\text{moles of nitrogen gas} = \frac{12.5 \, \text{g}}{28 \, \text{g/mol}} = 0.4464 \, \text{moles}
\][/tex]
3. Relate the moles of nitrogen gas to the moles of sodium azide:
According to the balanced chemical equation:
[tex]\[
2 \, \text{NaN}_3 \rightarrow 2 \, \text{Na} + 3 \, \text{N}_2
\][/tex]
From the equation, 2 moles of [tex]\(\text{NaN}_3\)[/tex] produce 3 moles of [tex]\(\text{N}_2\)[/tex]. Thus, 1 mole of [tex]\(\text{N}_2\)[/tex] requires:
[tex]\[
\frac{2 \, \text{moles of NaN}_3}{3 \, \text{moles of N}_2}
\][/tex]
To find the moles of [tex]\(\text{NaN}_3\)[/tex] required:
[tex]\[
\text{moles of NaN}_3 = \text{moles of N}_2 \times \frac{2}{3}
\][/tex]
[tex]\[
\text{moles of NaN}_3 = 0.4464 \, \text{moles} \times \frac{2}{3} = 0.2976 \, \text{moles}
\][/tex]
4. Convert the moles of sodium azide to grams:
The molar mass of sodium azide ([tex]\(\text{NaN}_3\)[/tex]) is 65 g/mol.
[tex]\[
\text{mass of NaN}_3 = \text{moles} \times \text{molar mass}
\][/tex]
[tex]\[
\text{mass of NaN}_3 = 0.2976 \, \text{moles} \times 65 \, \text{g/mol} = 19.35 \, \text{g}
\][/tex]
So, 19.35 grams of [tex]\( \text{NaN}_3 \)[/tex] are required to produce 10.0 liters of nitrogen gas under the given conditions.
