Question

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:

Answer 1

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 * $\frac{2molNaN_3}{3molN_2}$ = 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