To determine how many grams of gaseous nitrogen [tex]\(\left( \text{N}_2 \right)\)[/tex] are formed from 0.35 moles of sodium azide [tex]\(\left( \text{NaN}_3 \right)\)[/tex], follow these steps:
1. Understand the stoichiometry of the reaction:
The balanced chemical equation for the decomposition of sodium azide is:
[tex]\[
2 \text{NaN}_3(s) \rightarrow 2 \text{Na}(s) + 3 \text{N}_2(g)
\][/tex]
This equation indicates that 2 moles of sodium azide produce 3 moles of nitrogen gas.
2. Calculate the moles of nitrogen gas produced:
Given that the mole ratio between [tex]\(\text{NaN}_3\)[/tex] and [tex]\(\text{N}_2\)[/tex] is [tex]\(2:3\)[/tex], we can find the moles of [tex]\(\text{N}_2\)[/tex] produced from 0.35 moles of [tex]\(\text{NaN}_3\)[/tex]:
[tex]\[
\text{Moles of } \text{N}_2 = 0.35 \, \text{moles of } \text{NaN}_3 \times \frac{3 \, \text{moles of } \text{N}_2}{2 \, \text{moles of } \text{NaN}_3}
\][/tex]
[tex]\[
\text{Moles of } \text{N}_2 = 0.525 \, \text{moles}
\][/tex]
3. Determine the molar mass of nitrogen gas [tex]\(\left( \text{N}_2 \right)\)[/tex]:
Nitrogen gas is diatomic, so the molar mass of [tex]\(\text{N}_2\)[/tex] is:
[tex]\[
\text{Molar mass of } \text{N}_2 = 2 \times 14.01 \, \text{g/mol} = 28.02 \, \text{g/mol}
\][/tex]
4. Convert moles of nitrogen gas to grams:
Use the molar mass to convert the moles of [tex]\(\text{N}_2\)[/tex] to grams:
[tex]\[
\text{Mass of } \text{N}_2 = \text{Moles of } \text{N}_2 \times \text{Molar mass of } \text{N}_2
\][/tex]
[tex]\[
\text{Mass of } \text{N}_2 = 0.525 \, \text{moles} \times 28.02 \, \text{g/mol}
\][/tex]
[tex]\[
\text{Mass of } \text{N}_2 = 14.7105 \, \text{grams}
\][/tex]
5. Significant Figures:
The initial quantity of [tex]\(\text{NaN}_3\)[/tex] (0.35 moles) has 2 significant figures. Therefore, our final answer should also be reported to 2 significant figures:
[tex]\[
\text{Mass of } \text{N}_2 \approx 15 \, \text{g} \, \text{(rounded to 2 significant figures)}
\][/tex]
Hence, the mass of gaseous nitrogen [tex]\(\left( \text{N}_2 \right)\)[/tex] formed from 0.35 moles of sodium azide [tex]\(\left( \text{NaN}_3 \right)\)[/tex] is approximately [tex]\(14.71\)[/tex] grams.
