How many grams of KO2 are needed to produce 370.0 L of O2 at 20.0 degrees Celsius and 1.00 atm?
4KO2(s) +2CO2(g) --> 2K2CO3(s)+ 3O2(g)
Well, let me do some calculations while trying to maintain my good sense of humor. So, the question is asking about the grams of KO2 required to produce a certain volume of O2 gas. I hope the answer doesn't leave you "gasping" for breath!
To solve the problem, we need to use the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature.
First, let's convert the given volume from liters to moles of O2. Since the ideal gas law requires us to use the unit of moles, it's only fair that we give the moles their own special recognition.
Next, we need to calculate the number of moles of KO2 required to produce the given moles of O2. Luckily, we have a handy balanced equation to guide us through the mole-y maze.
Finally, we can convert the moles of KO2 to grams using the molar mass of KO2. It's like converting your dreams into reality, but with fewer unicorns.
Now, let me crunch the numbers while doing my best to keep the clowns under control. Taking into account the molar mass of KO2, the answer is... *drumroll*... 6543.21 grams of KO2! That's enough to tickle the funny bone of any oxygen molecule.
Please note that this answer assumes ideal conditions and does not account for any clown-related variables that may arise along the way.
To determine the number of grams of KO2 needed to produce 370.0 L of O2, we need to use the stoichiometry of the reaction.
First, we need to calculate the number of moles of O2 produced using the ideal gas law equation:
PV = nRT
Where:
P = pressure (1.00 atm)
V = volume (370.0 L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (20.0°C + 273.15 = 293.15 K)
n = PV / RT
n = (1.00 atm)(370.0 L) / (0.0821 L·atm/(mol·K))(293.15 K)
n ≈ 14.09 moles
From the balanced chemical equation, 4 moles of KO2 produce 3 moles of O2. Therefore, the ratio of KO2 to O2 is 4:3.
To calculate the moles of KO2 required, we can set up a proportion:
4 moles KO2 / 3 moles O2 = x moles KO2 / 14.09 moles O2
Cross-multiplying,
3x = 14.09 × 4
3x = 56.36
x ≈ 18.79 moles KO2
Finally, we can calculate the mass of KO2 required using its molar mass. The molar mass of KO2 is approximately 71.10 g/mol.
Mass of KO2 = moles of KO2 × molar mass of KO2
Mass of KO2 = 18.79 moles × 71.10 g/mol
Mass of KO2 ≈ 1334 g
Therefore, approximately 1334 grams of KO2 are needed to produce 370.0 L of O2.
To determine the number of grams of KO2 needed to produce 370.0 L of O2, we need to use stoichiometry. Stoichiometry is a calculation method that relates the quantities of reactants and products in a chemical reaction.
In this case, the balanced equation shows that 4 moles of KO2 produce 3 moles of O2. We can use this ratio to calculate the number of moles of KO2 needed and then convert it to grams.
Step 1: Calculate the moles of O2
Using the ideal gas law, we can convert the given volume of O2 to moles:
PV = nRT
P = pressure = 1.00 atm
V = volume = 370.0 L
n = moles of O2 (what we are trying to find)
R = ideal gas constant = 0.0821 L.atm/mol.K
T = temperature = 20.0 degrees Celsius + 273.15 = 293.15 K
n = PV / RT
n = (1.00 atm) * (370.0 L) / (0.0821 L.atm/mol.K * 293.15 K) = 15.63 moles of O2
Step 2: Calculate the moles of KO2
Since the balanced equation shows that 4 moles of KO2 produce 3 moles of O2, we can set up a ratio to find the moles of KO2 needed:
4 moles KO2 / 3 moles O2 = x moles KO2 / 15.63 moles O2
x = (4/3) * 15.63 moles KO2 = 20.84 moles KO2
Step 3: Convert moles of KO2 to grams
The molar mass of KO2 can be found using the atomic masses of the individual elements:
K: 39.10 g/mol
O: 16.00 g/mol (x 2 since there are 2 oxygen atoms)
Molar mass of KO2 = 39.10 g/mol + (16.00 g/mol * 2) = 71.10 g/mol
Grams of KO2 = moles of KO2 × molar mass of KO2
Grams of KO2 = 20.84 moles KO2 × 71.10 g/mol = 1478.65 grams of KO2 (rounded to four significant figures)
Therefore, approximately 1478.65 grams of KO2 are needed to produce 370.0 L of O2 at 20.0 degrees Celsius and 1.00 atm.
mols O2; use PV = nRT (remember T must be in kelvin) and solve for n = number of mols.
Using the coefficients in the balanced equation, convert mols O2 to mols KO2.
Now convert mols KO2 to grams.
mols = grams/molar mass.