How many liters of oxygen gas at 153 degrees celsius and 0.820 atm can be produced by the decomposition of 22.4 g of solid KClO3? (The other decomposition product is solid potassium chloride)
To determine the number of liters of oxygen gas produced, we need to use the ideal gas law equation, PV = nRT, where:
P is the pressure in atm (0.820 atm),
V is the volume in liters (unknown),
n is the number of moles of gas (unknown),
R is the ideal gas constant (0.0821 L*atm/(mol*K)),
T is the temperature in Kelvin.
1. First, we need to convert the temperature from Celsius to Kelvin.
The given temperature is 153 degrees Celsius.
To convert to Kelvin, we add 273.15 to the Celsius temperature:
T(K) = 153°C + 273.15 = 426.15 K.
2. Next, we need to calculate the number of moles of oxygen gas produced based on the given mass of KClO3.
The molar mass of KClO3 is the sum of the atomic masses of its elements, which can be found from the periodic table:
K = 39.10 g/mol (potassium),
Cl = 35.45 g/mol (chlorine),
O = 16.00 g/mol (oxygen).
Hence,
Molar mass of KClO3 = 39.10 g/mol + 35.45 g/mol + (3 × 16.00 g/mol) = 122.55 g/mol.
Given mass of KClO3 = 22.4 g.
To find the number of moles, we divide the mass by the molar mass:
n(KClO3) = 22.4 g / 122.55 g/mol = 0.183 mol.
3. The balanced chemical equation for the decomposition of KClO3 is:
KClO3(s) -> KCl(s) + 3/2 O2(g)
From the balanced equation, we know that for every 2 moles of KClO3, we get 3/2 moles of O2.
Therefore, the number of moles of O2 produced can be calculated as follows:
n(O2) = (3/2) * n(KClO3)
= (3/2) * 0.183 mol
= 0.2745 mol.
4. Now, we can use the ideal gas law to calculate the volume of the gas.
Rearranging the equation, we have:
V = (nRT) / P.
Substituting the values into the equation:
V = (0.2745 mol * 0.0821 L*atm/(mol*K) * 426.15 K) / 0.820 atm
Calculating this expression gives:
V = 11.359 L.
Therefore, approximately 11.359 liters of oxygen gas would be produced.
To determine the number of liters of oxygen gas produced by the decomposition of KClO3, we need to 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) = 153 °C + 273.15
T(K) = 426.15 K
Next, we need to find the number of moles of oxygen gas produced. We can use the molar mass of KClO3 to convert the given mass into moles.
Molar mass of KClO3:
K: 39.10 g/mol
Cl: 35.45 g/mol
O: 16.00 g/mol
Molar mass of KClO3 = (39.10 g/mol) + (35.45 g/mol) + (3 × 16.00 g/mol) = 122.55 g/mol
Number of moles of KClO3 = Mass of KClO3 / Molar mass of KClO3
Number of moles of KClO3 = 22.4 g / 122.55 g/mol = 0.1827 mol
According to the balanced chemical equation for the decomposition of KClO3:
2 KClO3 -> 2 KCl + 3 O2
The stoichiometry tells us that for every 2 moles of KClO3, we will produce 3 moles of O2.
Number of moles of O2 = (Number of moles of KClO3) x (3 moles of O2 / 2 moles of KClO3)
Number of moles of O2 = 0.1827 mol x (3/2) = 0.2741 mol
Now, we can use the ideal gas law to calculate the volume of oxygen gas.
PV = nRT
V = (nRT) / P
V = (0.2741 mol) × (0.0821 L·atm/mol·K) × (426.15 K) / 0.820 atm
V = 9.004 L
Therefore, approximately 9.004 liters of oxygen gas can be produced by the decomposition of 22.4 g of KClO3 at 153 degrees Celsius and 0.820 atm.
1. Write the equation and balance it.
2. Convert what you have (KClO3) to moles. # moles = grams/molar mass.
3. Using the coefficient in the balanced equation, convert mole KClO3 to moles O2.
4. Now use PV = nRT to convert moles O2 at STP to liters of O2 at the other conditions.
Post your work if you get stuck.