What volume of oxygen gas can be collected
at 0.565 atm pressure and 48.0◦C when 42.9g of KClO3 decompose by heating, according
to the following equation?
2 KClO3(s) ∆−−−−→2KCl(s) + 3O2(g)
MnO2
Sara, I worked this for you under your othr screen name of Anon above.
To calculate the volume of oxygen gas collected, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = moles of gas
R = ideal gas constant
T = temperature (in Kelvin)
First, let's convert the given temperature from Celsius to Kelvin using the equation:
T(K) = T(°C) + 273.15
T(K) = 48.0 + 273.15 = 321.15 K
Next, we need to calculate the number of moles of oxygen gas produced using stoichiometry. According to the balanced equation, for every 2 moles of KClO3, 3 moles of O2 are produced.
Step 1: Convert the given mass of KClO3 to moles.
42.9g KClO3 * (1 mol KClO3 / molar mass of KClO3)
Where the molar mass of KClO3 is:
K (39.10 g/mol) + Cl (35.45 g/mol) + 3O (16.00 g/mol) = 122.55 g/mol
42.9g KClO3 * (1 mol KClO3 / 122.55 g KClO3) = 0.350 mol KClO3
Step 2: Calculate the number of moles of O2 produced.
Using the stoichiometry, for every 2 moles of KClO3, 3 moles of O2 are produced.
0.350 mol KClO3 * (3 mol O2 / 2 mol KClO3) = 0.525 mol O2
Now we have all the values needed to calculate the volume of oxygen gas.
PV = nRT
V = (nRT) / P
V = (0.525 mol * 0.0821 atm·L/mol·K * 321.15 K) / 0.565 atm
Calculating the volume:
V = 9.87 L
Therefore, the volume of oxygen gas collected is approximately 9.87 liters.