Question

2. Sodium nitrate decomposes into sodium nitrite and oxygen gas with heat. When 0.123 g of sodium nitrate was heated, it produced 16.2mL of oxygen collected over water. The room temperature and pressure are 22.5ºC and 758torr, respectively.The water over which the gas was collected had a temperature of 20.0ºC. Calculate the theoretical amount of oxygen in grams which should have been produced. What is the percent yield?

Answer 1

Theoretical mass: 0.0232 g

Percent yield: 92.67%

Step-by-step 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%