how many grams of potassium chlorate decomposes to potassium chloride and 638 mL pf o2 at 128°C and 752 torr? 2KClO3(s) 2KCl(s) + 3O2(g)
To determine the mass of potassium chlorate that decomposes, we need to calculate the stoichiometry of the reaction. The balanced equation tells us that for every 2 moles of potassium chlorate that decompose, we obtain 3 moles of oxygen gas.
To calculate the number of moles of oxygen gas (O2), we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure in atm (convert 752 torr to atm)
V = volume in liters (convert 638 mL to liters)
n = number of moles of gas
R = ideal gas constant (0.0821 atm L/mol K)
T = temperature in Kelvin (add 273 to convert from °C)
So, let's plug in the values into the equation:
P = 752 torr * (1 atm / 760 torr) = 0.9895 atm
V = 638 mL * (1 L / 1000 mL) = 0.638 L
T = 128°C + 273 = 401 K
R = 0.0821 atm L/mol K
PV = nRT
(0.9895 atm)(0.638 L) = n(0.0821 atm L/mol K)(401 K)
Solving for n (number of moles):
n = (0.9895 atm * 0.638 L) / (0.0821 atm L/mol K * 401 K)
n ≈ 0.0156 moles of O2.
According to the stoichiometry in the balanced equation, for every 3 moles of O2, we get 2 moles of KClO3. Therefore, to obtain 0.0156 moles of O2, we need:
0.0156 moles O2 * (2 moles KClO3 / 3 moles O2) = 0.0104 moles KClO3.
Now, we can calculate the mass of potassium chlorate using its molar mass:
Mass = moles * molar mass
Molar mass of KClO3 = 122.55 g/mol (from periodic table)
Mass of KClO3 = 0.0104 moles * 122.55 g/mol
Mass of KClO3 ≈ 1.27 grams.
Therefore, approximately 1.27 grams of potassium chlorate decomposes to form potassium chloride (KCl) and 638 mL of oxygen gas (O2) at 128°C and 752 torr.