What volume of oxygen gas will be produced when 13.5 g of potassium chlorate is decomposed at STP?
2KClO3(s) → 2KCl(s) + 3 O2(g)
per the reaction equation ... each mole of KClO3 produces 1.5 moles of O2
13.5 g / (molar mass KClO3) = moles KClO3
... multiply by 1.5 to find moles of O2
each mole of O2 has a volume of 22.4 L
when 25 g of potassium chordate decomposes 15 g of paoassium chloride is produced. how many miles of oxygen has are produced
To determine the volume of oxygen gas produced, you can use the ideal gas law, which states:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal Gas Constant (0.0821 L·atm/mol·K)
T = Temperature (in Kelvin)
At STP (Standard Temperature and Pressure), the pressure is 1 atm, and the temperature is 273.15 K (0 degrees Celsius).
First, you need to determine the number of moles of potassium chlorate (KClO3) used in the reaction. To do this, you can use the molar mass of KClO3.
Molar mass of KClO3 = (1 * atomic mass of K) + (1 * atomic mass of Cl) + (3 * atomic mass of O)
= (1 * 39.10 g/mol) + (1 * 35.45 g/mol) + (3 * 16.00 g/mol)
= 122.55 g/mol
Next, you can calculate the number of moles of KClO3:
moles of KClO3 = mass / molar mass
= 13.5 g / 122.55 g/mol
= 0.1102 mol (rounded to four decimal places)
According to the balanced equation, 2 moles of KClO3 produce 3 moles of O2.
So, using the mole ratio, the number of moles of O2 produced can be calculated as:
moles of O2 = (moles of KClO3) * (3 moles of O2 / 2 moles of KClO3)
= 0.1102 mol * (3/2)
= 0.1653 mol (rounded to four decimal places)
Now, you can plug all the values into the ideal gas law equation to find the volume of oxygen gas produced:
PV = nRT
V = (nRT) / P
= (0.1653 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm
= 3.615 L (rounded to three decimal places)
Therefore, 13.5 g of potassium chlorate decomposed at STP will produce approximately 3.615 liters of oxygen gas.