Answer:
Theoretical mass: 0.0232 g
Percent yield: 92.67%
Explanation:
Sodium nitrate is NaNO₃, and sodium nitrite is NaNO₂, the decomposes occurs by the reaction:
2NaNO₃(s) → 2NaNO₂(s) + O₂(g)
The molar masses are: Na = 23 g/mol, N = 14 g/mol, O = 16 g/mol.
NaNO₃ = 23 + 14 + 3x16 = 85 g/mol
So the number of moles of NaNO₃ is:
n = 0.123g/85 g/mol = 0.00145 mol
The theoretical amount of oxygen is given by the reaction stoichiometry:
2 mol of NaNO₃ ------------- 1 mol of O₂
0.00145 mol -------------- x
By a simple direct three rule:
2x = 0.00145
x = 7.25x10⁻⁴ mol of O₂
The theoretical mass is:
m = 32 g/mol *7.25x10⁻⁴ mol
m = 0.0232 g
The real mass can be calculated by the ideal gas equation:
PV = nRT,
Where P is the pressure (758 torr = 0.997 atm), V is the volume (16.2 mL = 0.0162 L), n is the number of moles, R is the gas constant ( 0.082 atm*L/mol*K), and T is the temperature (20.0ºC = 293 K).
The volume was measured in water, so we must use the temperature at that system:
0.997*0.0162 = n*0.082*293
24.026n = 0.0161514
n = 6.72x10⁻⁴ mol
So the mass of the oxygen collected is (mc):
mc = 32 g/mol *6.72x10⁻⁴ mol
mc = 0.0215 g
The yield(y) is the mass collected divideb by the theoretical mass:
y = 0.0215/0.0232
y = 0.9267
y = 92.67%
