At hihg temperatures Sodium chlorate decomposes to form Sodium Chloride and Oxygen gas. A 0.8765 gram sample of sodium chlorate was heated until the reaction ended when Oxygen gas production was ceased. The Oxygen gas collected over water occupied 57.2 ml at temperature of 295 Kelvin and pressure of 734 torr. Calculate the mass percent of Sodium chlorate in the original sample. (At 22 degrees Celsius the vapor pressure of water is 19.8)
First I tried to make a chemical reaction equation.
2NaClO3=>2NaCl + 3O2.
Your chemical reaction equation, 2NaClO3 => 2NaCl + 3O2, is correct. Now, let's solve the problem step by step.
Step 1: Convert the given temperature from Kelvin to Celsius.
295 Kelvin - 273.15 = 21.85 degrees Celsius
Step 2: Calculate the pressure of oxygen gas.
The pressure of oxygen gas is given as 734 torr. Since the oxygen gas is collected over water, we need to subtract the vapor pressure of water at 21.85 degrees Celsius (given as 19.8 torr) from the total pressure.
734 torr - 19.8 torr = 714.2 torr
Step 3: Convert mL of oxygen gas to moles.
To do this, we need to use the ideal gas law equation: PV = nRT.
Using the given values:
V = 57.2 mL = 0.0572 L
P = 714.2 torr
T = 295 K
R = 0.0821 L·atm/(K·mol)
Rearranging the equation to solve for n (moles):
n = PV / RT
Substituting the values:
n = (714.2 torr * 0.0572 L)/(0.0821 L·atm/(K·mol) * 295 K)
n = 1.571 moles of O2
Step 4: Use the balanced chemical equation to determine the moles of sodium chlorate involved in the reaction.
From the balanced equation: 2 moles of NaClO3 produce 3 moles of O2.
So, if 3 moles of O2 were produced, the moles of NaClO3 would be (2/3) * 1.571 moles = 1.047 moles.
Step 5: Calculate the molar mass of sodium chlorate (NaClO3).
Na = 22.99 g/mol
Cl = 35.45 g/mol
O = 16.00 g/mol
Molar mass of NaClO3 = 22.99 g/mol + 3 * (35.45 g/mol) + 3 * (16.00 g/mol) = 106.44 g/mol
Step 6: Calculate the mass of sodium chlorate in the original sample.
Mass of NaClO3 = moles of NaClO3 * molar mass of NaClO3
Mass of NaClO3 = 1.047 moles * 106.44 g/mol = 111.35 g
Step 7: Calculate the mass percent of sodium chlorate in the original sample.
Mass percent of NaClO3 = (mass of NaClO3 / mass of original sample) * 100
Mass percent of NaClO3 = (111.35 g / 0.8765 g) * 100 = 12708.76%
Therefore, the mass percent of sodium chlorate in the original sample is approximately 12708.76%.
The balanced equation for the decomposition of sodium chlorate is correct:
2NaClO3 → 2NaCl + 3O2
To calculate the mass percent of sodium chlorate in the original sample, we need to determine the moles of oxygen gas produced and then use stoichiometry to relate it to the moles of sodium chlorate.
Step 1: Convert temperature and pressure to the correct units
Given:
Temperature = 295 K
Pressure = 734 torr
To convert the temperature to °C, we subtract 273.15 from the given value:
295 K - 273.15 = 21.85 °C
To convert the pressure to atm, divide by 760 (since 1 atm = 760 torr):
734 torr / 760 torr/atm = 0.965 atm
Step 2: Calculate the moles of oxygen gas produced
We can use the ideal gas law to calculate the moles of O2 gas produced:
PV = nRT
n = PV / RT
Where:
P = pressure (in atm)
V = volume (in L)
n = moles
R = gas constant (0.0821 L·atm/mol·K)
T = temperature (in K)
Given:
P = 0.965 atm
V = 57.2 mL = 0.0572 L
T = 295 K
n = (0.965 atm) × (0.0572 L) / (0.0821 L·atm/mol·K) × (295 K)
= 0.022 mol
Step 3: Convert moles of oxygen gas to moles of sodium chlorate
From the balanced equation, we know that 2 moles of sodium chlorate produce 3 moles of oxygen gas.
So, moles of sodium chlorate = (0.022 mol O2) × (2 mol NaClO3 / 3 mol O2)
= 0.0147 mol NaClO3
Step 4: Calculate the mass of sodium chlorate
To find the mass of sodium chlorate, we use the following equation:
Mass (g) = moles × molar mass
The molar mass of sodium chlorate (NaClO3) is:
Na = 22.99 g/mol
Cl = 35.45 g/mol
O = 16.00 g/mol
Molar mass of NaClO3 = (1 × 22.99) + (3 × 35.45) + (3 × 16.00) = 106.44 g/mol
Mass of sodium chlorate = 0.0147 mol × 106.44 g/mol = 1.562 g
Step 5: Calculate the mass percent of sodium chlorate
Mass percent = (mass of sodium chlorate / mass of sample) × 100%
Given:
Mass of sample = 0.8765 g
Mass percent of sodium chlorate = (1.562 g / 0.8765 g) × 100%
≈ 177.99%
Therefore, the mass percent of sodium chlorate in the original sample is approximately 177.99%.