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) ∆−→MnO2 2KCl(s) + 3O2(g)
Abby, Anon, Sarah, et al. You certainly make things difficult by not sticking to the same screen name. I've answered this above.
To find the volume of oxygen gas collected, we can use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure of the gas (in atmospheres)
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 Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 48.0 + 273.15
T(K) = 321.15 K
Next, we need to determine the number of moles of oxygen gas produced by the decomposition of KClO3. From the balanced equation:
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
It shows that for every 2 moles of KClO3, 3 moles of O2 are produced.
First, we need to calculate the number of moles of KClO3 used:
m(KClO3) = 42.9 g
The molar mass of KClO3 can be calculated as follows:
M(KClO3) = M(K) + M(Cl) + 3 * M(O)
M(KClO3) = 39.10 g/mol + 35.45 g/mol + 3 * 16.00 g/mol
M(KClO3) = 122.55 g/mol
The number of moles of KClO3 can now be determined:
n(KClO3) = m(KClO3) / M(KClO3)
n(KClO3) = 42.9 g / 122.55 g/mol
n(KClO3) ≈ 0.350 mol
From the stoichiometry of the equation, we know that for every 2 moles of KClO3, 3 moles of O2 are produced. Therefore, the number of moles of O2 can be calculated as follows:
n(O2) = (3/2) * n(KClO3)
n(O2) = (3/2) * 0.350 mol
n(O2) = 0.525 mol
Now that we have the number of moles of O2 and the other values needed for the Ideal Gas Law equation, we can calculate the volume of the gas:
PV = nRT
V = (nRT) / P
V = (0.525 mol * 0.0821 L.atm/mol.K * 321.15 K) / 0.565 atm
V ≈ 24.18 L
Therefore, approximately 24.18 liters of oxygen gas can be collected at a pressure of 0.565 atm and a temperature of 48.0°C when 42.9 g of KClO3 decompose.