What volume of oxygen gas can be collected
at 1.26 atm pressure and 2.00◦C when 39.7 g
of KClO3 decompose by heating, according to
the following equation?
2 KClO3(s) ∆/MnO2 −−−−→ 2 KCl(s) + 3 O2(g)
Answer in units of L.
To calculate the volume of oxygen gas (O2) collected, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, let's convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 2.00 + 273.15
T(K) = 275.15 K
Next, let's calculate the number of moles of oxygen gas (O2) produced using the stoichiometry of the balanced equation.
According to the equation: 2 KClO3(s) → 3 O2(g)
The molar mass of KClO3 is:
M(KClO3) = (39.10 g/mol * 2) + 35.45 g/mol + 16.00 g/mol * 3 = 122.55 g/mol
Now we can calculate the number of moles of KClO3 used:
n(KClO3) = Mass(KClO3) / Molar mass(KClO3)
n(KClO3) = 39.7 g / 122.55 g/mol
Next, using the stoichiometry of the balanced equation, we can determine the number of moles of O2 produced:
n(O2) = 3 * n(KClO3)
Now we can use the ideal gas law equation to calculate the volume of O2 gas collected:
V = nRT / P
V = (n(O2) * R * T) / P
Substituting the values into the equation:
V = (n(O2) * R * T) / P
V = (3 * n(KClO3) * R * T) / P
Calculate the volume of O2 gas collected.
To solve this problem, we need to use the ideal gas law formula which is PV = nRT, where:
P = pressure of the gas (in atm)
V = volume of the gas (in liters)
n = number of moles of the gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature of the gas (in Kelvin)
First, we need to convert the given temperature from °C to Kelvin. The equation for that is:
T(°C) + 273.15 = T(K)
Given temperature: 2.00°C + 273.15 = 275.15 K
Next, we need to determine the number of moles of oxygen gas (O2) using stoichiometry:
From the balanced equation, we can see that for every 2 moles of KClO3 decomposed, 3 moles of O2 gas are produced.
Since we have 39.7 g of KClO3, we can calculate the number of moles using its molar mass:
Molar mass of KClO3 = (39.10 g/mol K) + (35.45 g/mol Cl) + (3 x 16.00 g/mol O)
Molar mass of KClO3 = 122.55 g/mol
Number of moles of KClO3 = 39.7 g / 122.55 g/mol ≈ 0.3244 mol
Using the stoichiometric ratio, we can determine the number of moles of O2 gas produced:
Number of moles of O2 = (3 mol O2 / 2 mol KClO3) x (0.3244 mol KClO3)
Number of moles of O2 = 0.4866 mol
Now, we have all the information needed to solve for the volume of oxygen gas (V) using the ideal gas law:
PV = nRT
Substituting the given values:
P = 1.26 atm
V = ?
n = 0.4866 mol
R = 0.0821 L·atm/mol·K
T = 275.15 K
Rearranging the equation to solve for V:
V = (nRT) / P
V = (0.4866 mol)(0.0821 L·atm/mol·K)(275.15 K) / 1.26 atm
Calculating the value:
V ≈ 10.643 L
Therefore, the volume of oxygen gas collected under the given conditions is approximately 10.643 liters.
See your other post above but this one is not eligible for the shortcut.
Convert g KClO3 to mols. mols = g/molar mass = ?
Then convert mols KClO3 to mols O2
Use PV = nRT and convert mols CO2 to L CO2 at the conditions listed.