Question 6.10: Automotive airbags inflate when sodium azide, NaN3, rapidly ......

Automotive airbags inflate when sodium azide, NaN3NaN_{3}, rapidly decomposes to its constituent elements. The equation for the chemical reaction is
2NaN3(s)2Na(s)+3N2(g)\quad\quad\quad\quad 2NaN_{3}(s) → 2Na(s) + 3N_{2}(g)
The gaseous N2N_2 so generated inflates the airbag (see Figure 6.10). How many moles of NaN3NaN_{3} would have to decompose in order to generate 253 million (2.53 × 10810^{8}) molecules of N2N_2?

6.10
Step-by-Step
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Although a calculation of this type does not have a lot of practical significance, it tests your understanding of the problem-solving relationships discussed in this section of the text.
Step 1: The given quantity is 2.53 × 10810^{8} molecules of N2N_2, and the desired quantity is moles of NaN3NaN_{3}.
\quad\quad\quad\quad2.53 × 10810^{8} molecules N2N_2 = ? moles NaN3NaN_{3}
In terms of Figure 6.9, this is a “particles of A” to “moles of B” problem.
Step 2: Using Figure 6.9 as a road map, we determine that the pathway for this problem is

Particles of AnumberAvogadrosMoles of AcoefficientsEquationMoles of B\quad\quad\quad \boxed{Particles  of  A} \xrightarrow[number]{Avogadro’s} \boxed{Moles  of  A} \xrightarrow[coefficients]{Equation} \boxed{Moles  of  B}

The mathematical setup is

2.53×108 molecules N2×(1 mole N26.02×1023 molecules N2)×(2 moles NaN33 moles N2)\quad\quad 2.53\times 10^{8}  \cancel{molecules  N_{2}}\times (\frac{1  \cancel{mole  N_{2}}}{6.02\times 10^{23}  \cancel{molecules  N_{2}}})\times (\frac{2  \cancel{moles  NaN_{3}}}{3  \cancel{moles  N_{2}}})

Avogadro’s number is present in the first conversion factor. The 2 and 3 in the second conversion factor are the coefficients, respectively, of NaN3NaN_{3} and N2N_2 in the balanced chemical equation.
Step 3: The solution to the problem, obtained by doing the arithmetic after all the numerical factors have been collected, is

(2.53×108×1×26.02×1023×3) mole NaN3=2.80×1016 mole NaN3\quad\quad\quad (\frac{2.53\times 10^{8}\times 1\times 2}{6.02\times 10^{23}\times 3})  mole  NaN_{3}=2.80\times 10^{-16}  mole  NaN_{3}
6.9

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