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Chapter 5: Problem 101

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 \(\mathrm{mL}\) at a temperature of $22^{\circ} \mathrm{C}$ and a pressure of 734 torr. Calculate the mass percent of \(\mathrm{NaClO}_{3}\) in the original sample. (At \(22^{\circ} \mathrm{C}\) the vapor pressure of water is 19.8 torr.)

Short Answer

Expert verified
The mass percent of $\mathrm{NaClO}_{3}$ in the original sample is approximately 21.892%.

Step by step solution

01

Calculate the partial pressure of oxygen gas

To find the partial pressure of oxygen gas, we subtract the vapor pressure of water at 22°C (19.8 torr) from the total pressure exerted on the gas mixture (734 torr). This will give us the partial pressure of oxygen gas. Partial pressure of O2 = Total pressure - Water vapor pressure Partial pressure of O2 = 734 torr - 19.8 torr = 714.2 torr
02

Use the Ideal Gas Law to determine the moles of oxygen gas

Next, we will use the Ideal Gas Law, which is given by: PV = nRT Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. First, convert the given temperature to Kelvin: T = 22°C + 273.15 = 295.15 K Now convert the volume from mL to L: V = 57.2 mL × (1 L / 1,000 mL) = 0.0572 L Use the Ideal Gas Law with the constant R = 0.08206 (L atm) / (mol K) to find the moles of oxygen gas, we also need to convert the pressure from torr to atm: P = 714.2 torr × (1 atm / 760 torr) = 0.940263 atm n = PV / RT n (O2) = (0.940263 atm) × (0.0572 L) / ((0.08206 (L atm) / (mol K)) × (295.15 K)) n (O2) = 0.002707 mol
03

Convert the moles of oxygen gas to moles of sodium chlorate

The balanced decomposition reaction of sodium chlorate is: 2 NaClO3 -> 2 NaCl + 3 O2 From the balanced equation, 2 moles of sodium chlorate produce 3 moles of oxygen gas. To find the moles of sodium chlorate, we can use the stoichiometric ratio: n (NaClO3) = n (O2) × (2 / 3) n (NaClO3) = 0.002707 mol × (2 / 3) = 0.001804 mol
04

Determine the mass of pure sodium chlorate in the original sample

To find the mass of pure sodium chlorate in the original sample, we multiply the moles of sodium chlorate by its molar mass. Molar mass of NaClO3 = 22.99 g/mol (Na) + 35.45 g/mol (Cl) + 3 × (16.00 g/mol) (O) = 106.44 g/mol mass (NaClO3) = n (NaClO3) × Molar mass mass (NaClO3) = 0.001804 mol × 106.44 g/mol = 0.192026 g
05

Calculate the mass percent of sodium chlorate in the original sample

Finally, to determine the mass percent of sodium chlorate in the original sample, divide the mass of pure sodium chlorate by the mass of the original sample and then multiply by 100. mass percent (NaClO3) = (mass (NaClO3) / mass (original sample)) × 100 mass percent (NaClO3) = (0.192026 g / 0.8765 g) × 100 = 21.892 %

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