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?

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%

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol