The decomposition of KClO3 produces oxygen gas and solid KCl. What mass of KClO3 must be decomposed to produce 126 L of oxygen gas at 133 Celsius and .88 atm?
I know I need to use PV=nRT but I can't figure out how.
Here is a worked example on the decomposition of KClO3. To convert 126 L O2 to n = mols, use PV = nRT and solve for n, then follow the instructions beginning with step 3.
http://www.jiskha.com/science/chemistry/stoichiometry.html
thank you!
To calculate the mass of KClO3 that must be decomposed to produce a certain volume of oxygen gas, we can use the ideal gas law equation: PV = nRT. Since we are given the volume (V), temperature (T), and pressure (P) of the oxygen gas, we can rearrange the equation to solve for the number of moles (n) of oxygen gas produced.
First, convert the temperature from Celsius to Kelvin by adding 273.15:
133 Celsius = 133 + 273.15 = 406.15 K
Next, rearrange the equation to solve for n:
n = PV/RT
Plug in the values for pressure (P), volume (V), and temperature (T) into the equation:
n = (0.88 atm) * (126 L) / [(0.0821 L·atm/mol·K) * (406.15 K)]
Calculate the value for n:
n = 0.01147 mol
From the balanced chemical equation, we know that 2 moles of KClO3 produce 3 moles of oxygen gas.
Therefore, we can set up a ratio to find out the number of moles of KClO3 required to produce 0.01147 mol of oxygen gas:
2 moles KClO3 : 3 moles O2 = x moles KClO3 : 0.01147 mol O2
Solve for x:
x = (0.01147 mol O2 * 2 moles KClO3) / 3 moles O2
Calculate the value for x:
x = 0.00765 mol KClO3
Finally, calculate the mass of KClO3:
molar mass of KClO3 = 39.1 g/mol (K) + 35.45 g/mol (Cl) + 16 g/mol (O) * 3
mass of KClO3 = 0.00765 mol KClO3 * 122.55 g/mol (KClO3)
So the mass of KClO3 that must be decomposed is approximately 0.94 grams.
To solve this problem, you can use the Ideal Gas Law equation, PV = nRT, where P represents pressure, V represents volume, n represents the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's convert the given temperature from Celsius to Kelvin by adding 273.15:
T = 133 °C + 273.15 = 406.15 K
Next, rearrange the Ideal Gas Law equation to solve for the number of moles (n):
n = PV / RT
Now, let's substitute the given values into the equation:
P = 0.88 atm
V = 126 L
R = 0.0821 L·atm/(mol·K)
T = 406.15 K
n = (0.88 atm * 126 L) / (0.0821 L·atm/(mol·K) * 406.15 K)
Now, we can calculate the number of moles of oxygen gas produced.
n = 4.66 moles
By stoichiometry, we know that for every one mole of KClO3 decomposed, one mole of oxygen gas is produced. Therefore, the number of moles of KClO3 decomposed is also equal to 4.66 moles.
To find the mass of KClO3, you need to know its molar mass. The molar mass of KClO3 can be calculated as follows:
Molar mass of KClO3 = (molar mass of K) + (molar mass of Cl) + 3*(molar mass of O)
Using the atomic masses of potassium (K = 39.1 g/mol), chlorine (Cl = 35.5 g/mol), and oxygen (O = 16.0 g/mol):
Molar mass of KClO3 = (39.1 g/mol) + (35.5 g/mol) +3*(16.0 g/mol) = 122.6 g/mol
Finally, we can calculate the mass of KClO3 using the number of moles and the molar mass:
Mass of KClO3 = n * Molar mass of KClO3
Mass of KClO3 = 4.66 moles * 122.6 g/mol
Therefore, the mass of KClO3 required to produce 126 L of oxygen gas at 133 °C and 0.88 atm pressure is approximately 571.76 grams.