Answer:
141.0 g NaN₃.
Explanation:
It is a stichiometric problem.
Firstly, we should wright the reaction as a balanced equation:
2NaN₃ → 2Na + 3N₂
It is clear that 2.0 moles of NaN₃ decompose to 2.0 mole of Na and 3.0 moles of N₂.
Then we should convert the volume of N₂ (70.0 L) to moles via using the gas law of ideal gas: PV = nRT, n = PV / RT,
Where, P is the pressure of the gas in atm (P = 1.2 atm).
V is the volume of the gas in L(V = 70.0 L).
R is the general gas constant (R = 0.082 L.atm/mol.K).
T is the temperature in K(T = 315 K).
∴ n of N₂ = PV / RT = (1.2 atm) (70.0 L) / (0.082 L.atm/mol.K) (315 K) = 3.252 mole.
From the stichiometry:
2.0 moles of NaN₃ decomposes to → 3.0 moles of N₂
??? mole of NaN₃ decomposes to → 3.2520 moles of N₂
The number of moles of NaN₃ = (2.0 moles of NaN₃) (3.2520 moles of N₂) / (3.0 moles of N₂) = 2.168 mole.
Finally, we can convert the number of moles of NaN₃ to mass using the relation: m = n x molar mass.
Molar mass of NaN₃ = 65.0 g/mol.
the mass of NaN₃ = n x molar mass = (2.168 mole) (65.0 g/mol) = 140.92 g ≅ 141.0 g.
