Answer:
he balanced chemical equation for the reaction between NaN3 and Na2O is:
2 NaN3(s) → 2 Na(s) + 3 N2(g)
According to the stoichiometry of this equation, 2 moles of NaN3 will produce 2 moles of Na, which in turn will react with 3 moles of N2. Therefore, the volume of N2 gas produced is proportional to the volume of NaN3 used.
To find the volume of NaN3 required to produce 14.7 liters of N2, we need to use the ideal gas law:
PV = nRT
where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the gas constant, and T is the temperature of the gas.
Assuming standard temperature and pressure (STP), which is 0°C and 1 atmosphere, we can simplify the equation to:
V = n/22.4
where V is the volume of the gas in liters and n is the number of moles of the gas.
We can use this equation to convert the volume of N2 to moles:
n = PV/RT = (1 atm)(14.7 L)/(0.08206 L·atm/mol·K)(273 K) = 0.608 mol
According to the stoichiometry of the balanced equation, 2 moles of NaN3 will produce 0.608 mol of N2. Therefore, the number of moles of NaN3 required is:
n(NaN3) = 2 × n(N2) = 2 × 0.608 mol = 1.216 mol
Finally, we can use the molar volume of a gas at STP to convert the number of moles to volume:
V(NaN3) = n(NaN3)/22.4 = 1.216 mol/22.4 L/mol = 0.054 L
Therefore, 0.054 liters of NaN3 are required to produce 14.7 liters of Na2O.