What volume of oxygen gas can be collected
at 0.565 atm pressure and 48.0◦C when 42.9
g of KClO3 decompose by heating, according
to the following equation?
2 KClO3(s) ∆−−−−→2KCl(s) + 3O2(g)
MnO2
mols KClO3 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols KClO3 to mols O2.
Now use PV = nRT at the conditions listed to calculate volume in L O2.
I can get to the PV=nRT part. What number would you put for R?
You should have told me you can get to that and we could have cut out half of the work. If you use atm for P, then R is 0.08206 L*atm/mol*K.
What I was saying in the response was that with your help I got to that part, but I was still confused on the formula. I got it now. Thank you.
You're welcome. I'm sorry I misunderstood.
To find the volume of oxygen gas produced, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure of the gas (in atm)
V = volume of the gas (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature of the gas (in Kelvin)
We have the pressure (0.565 atm) and temperature (48.0°C), but we need to convert the temperature to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 48.0 + 273.15
T(K) ≈ 321.15 K
Next, we need to find the number of moles of oxygen gas produced. To do this, we need to determine the moles of KClO3 reacted and use the stoichiometry of the balanced equation.
Molar mass of KClO3:
K = 39.10 g/mol
Cl = 35.45 g/mol
O = 16.00 g/mol
KClO3 = (39.10 + 35.45 + (16.00 x 3)) g/mol
KClO3 ≈ 122.55 g/mol
Moles of KClO3:
moles = mass / molar mass
moles = 42.9 g / 122.55 g/mol
moles ≈ 0.350 mol
According to the balanced equation, 2 moles of KClO3 produce 3 moles of O2. Therefore, we can calculate the moles of O2 produced:
moles of O2 = (moles of KClO3) x (3 moles O2 / 2 moles KClO3)
moles of O2 = 0.350 mol x (3/2)
moles of O2 = 0.525 mol
Finally, we can rearrange the ideal gas law equation to solve for V:
V = (nRT) / P
V = (0.525 mol x 0.0821 L·atm/(mol·K) x 321.15 K) / 0.565 atm
V ≈ 24.73 L
Therefore, the volume of oxygen gas produced is approximately 24.73 liters.