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What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:
2 NaN3 is) 2Na (s) + 3N2 (8)

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maacastrobr

Answer:

108.6 g

Explanation:

  • 2NaN₃(s) → 2Na(s) + 3N₂(g)

First we use the PV=nRT formula to calculate the number of nitrogen moles:

  • P = 1.00 atm
  • V = 56.0 L
  • n = ?
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = 0 °C ⇒ 0 + 273.2 = 273.2 K

Inputting the data:

  • 1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K
  • n = 2.5 mol

Then we convert 2.5 moles of N₂ into moles of NaN₃, using the stoichiometric coefficients of the balanced reaction:

  • 2.5 mol N₂ * [tex]\frac{2molNaN_3}{3molN_2}[/tex] = 1.67 mol NaN₃

Finally we convert 1.67 moles of NaN₃ into grams, using its molar mass:

  • 1.67 mol * 65 g/mol = 108.6 g

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