Question

8. What mass of NaN3 is needed to inflate an 18.0 L air bag in an automobile at 25.0°C and 1.00 atm according to the following reaction? 6NaN3(s) + Fe2O3(s) + 3Na2O(s) + 2Fe(s) + 9N2(g)

Answer 1

Volume of the air bag (V) = 18.0 L

Temperature (T) = (273 + 25) K = 298 K

Pressure (P) = 1.00 atm

Let us say that the moles of N₂ gas is formed = n

Universal gas constant (R) = 0.0821 L.atm/mol.K

From ideal gas law,

PV = nRT

or, 1.00 atm x 18.0 L = n x 0.0821 L.atm/mol.K x 298 K

or, n = 0.736 mol

Now, from the balanced equation given, 9 mol of N₂ gas is formed from 6 mol of NaN₃

Therefore, 0.736 mol of N₂ gas is formed from = (6 mol x 0.736 mol)/9 mol = 0.491 mol of NaN₃

Molar mass of NaN₃ = 65.0 g/mol

Hence, the mass of NaN₃ needed = 0.491 mol x 65.0 g/mol = 31.9 g