Question

At elevated temperatures, sodium chlorate decomposes to produce sodium chloride and oxygen gas. A 0.8765-g sample of impure sodium chlorate was heated until the production of oxygen gas ceased. The oxygen gas collected over water occupied 57.2 mL at a temperature of 22ºC and a pressure of 734 torr. Calculate the mass percent of NaHCO₃ in the original sample. (At 22ºC the vapor pressure of water is 19.8 torr.)

Answer 1

The mass percent of sodium chlorate in the original sample is calculated using the ideal gas law and stoichiometry, after correcting for the vapor pressure of water; it is found to be approximately 17.75%.

Step-by-step explanation:

The question asks us to calculate the mass percent of sodium chlorate ( NaClO₃ ) in an impure sample, given the mass of the sample and the volume of oxygen gas collected after decomposition. To find the mass percent, we first use the ideal gas law to determine the number of moles of oxygen produced, then apply stoichiometry to find the moles of NaClO₃ that decomposed, convert this to mass, and finally calculate the mass percent.

Firstly, correct the gas pressure of oxygen by subtracting the vapor pressure of water: 734 torr - 19.8 torr = 714.2 torr.

Using the ideal gas law ( PV = nRT ), where P is pressure in atm (1 torr = 1/760 atm), V is volume in liters, n is the number of moles, R is the gas constant (0.0821 L atm/mol K), and T is temperature in Kelvin (22°C + 273 = 295 K).

Convert volume from mL to L: 57.2 mL = 0.0572 L.

Convert pressure from torr to atm: 714.2 torr * (1 atm / 760 torr) = 0.940 atm.

Calculate moles of O₂ ( n ): n = PV / RT = (0.940 atm) * (0.0572 L) / (0.0821 L atm/mol K * 295 K) ≈ 0.00219 moles of O₂.

The decomposition reaction of sodium chlorate is 2NaClO₃ (s) → 2NaCl (s) + 3O₂ (g). According to stoichiometry, 3 moles of O₂ come from 2 moles of NaClO₃, therefore, 0.00219 moles of O₂ come from (0.00219 moles O₂ * 2 moles NaClO₃ / 3 moles O₂) ≈ 0.00146 moles of NaClO₃.

Convert moles of NaClO₃ to mass: m = n * M, where M is molar mass (NaClO₃ = 106.44 g/mol). m ≈ 0.00146 moles * 106.44 g/mol ≈ 0.1556 g of NaClO₃.

Finally, calculate the mass percent of NaClO₃: (mass of NaClO₃ / mass of sample) * 100% = (0.1556 g / 0.8765 g) * 100% ≈ 17.75%.

Answer 2

z = 18,46%

Step-by-step explanation:

First we need to discover the mole number of the O2 produced so we use PV=nRT

734 torr * 57.2*10^-3 L = n 62.3637 * 295.15 K

n =
`(734 * 57.2*10^-3)/(62.3637 * 295.15)`

n =
`(41,98)/(18.406)`

n = 2,28 * 10^-3 mole

The reaction to the decomposition is:

2 NaClO3 --> 2 NaCl + 3 O2

2 mole NaClO3 - 3 moles O2

x - 2,28*10^-3 moles O2

X= 1,52*10^-3 moles of NaClO3

1 moles of NaClO3 - 106,44 g

1,52*10^-3 moles of NaClO3 - y

y = 0,1617 g

0,8765 g - 100%

0,1617 g - z

z = 18,46%