The decomposition of potassium chlorate, KClO3, gives oxygen gas and solid potassium chloride. How many grams of KClO3 are needed to produce 5.00 L of Oxygen gas at STP?
Suppose that the reaction described above produces 4.80 L of O2 in the laboratory. What is the percent yield?
Here is a worked example using KClO3.
Remember mols x 22.4 L/mol = L.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the grams of KClO3 needed to produce 5.00 L of oxygen gas at STP, we will calculate the mole ratio between KClO3 and O2 using the balanced equation.
The balanced equation for the decomposition of KClO3 is:
2 KClO3 -> 2 KCl + 3 O2
From the equation, we can see that 2 moles of KClO3 will produce 3 moles of O2.
Step 1: Convert liters of O2 gas to moles.
At STP (standard temperature and pressure), 1 mole of any ideal gas occupies 22.4 L. So, to convert 5.00 L of O2 gas to moles, we use the ratio:
1 mole O2 / 22.4 L O2 = x moles O2 / 5.00 L O2
x = (5.00 L O2)(1 mole O2 / 22.4 L O2)
x = 0.2232 moles O2
Step 2: Calculate the moles of KClO3 needed.
According to the mole ratio from the balanced equation, we know that 2 moles of KClO3 produce 3 moles of O2.
Therefore, the moles of KClO3 needed can be calculated using the ratio:
2 moles KClO3 / 3 moles O2 = y moles KClO3 / 0.2232 moles O2
y = (0.2232 moles O2)(2 moles KClO3 / 3 moles O2)
y = 0.1488 moles KClO3
Step 3: Convert moles of KClO3 to grams.
To convert moles of KClO3 to grams, we need to know its molar mass. The molar mass of KClO3 is composed of the atomic masses of each element:
Molar mass KClO3 = (atomic mass of K) + (atomic mass of Cl) + 3 x (atomic mass of O)
= (39.10 g/mol) + (35.45 g/mol) + 3 x (16.00 g/mol)
= 122.55 g/mol
Now we can convert moles of KClO3 to grams:
Grams KClO3 = 0.1488 moles KClO3 x (122.55 g KClO3 / 1 mole KClO3)
Grams KClO3 = 18.22 g
Therefore, approximately 18.22 grams of KClO3 are needed to produce 5.00 L of oxygen gas at STP.
For the second part of the question, to calculate the percent yield, we need the actual yield and the theoretical yield of oxygen gas.
Theoretical yield: The maximum amount of oxygen gas that can be produced according to the balanced equation. In this case, it is 4.80 L.
Percent yield = (Actual yield / Theoretical yield) x 100
Percent yield = (4.80 L / 5.00 L) x 100
Percent yield = 96%
Therefore, the percent yield of the reaction is 96%.
To calculate the grams of KClO3 needed to produce 5.00 L of oxygen gas at STP, we need to use stoichiometry.
First, let's write the balanced equation for the reaction:
2 KClO3 --> 2 KCl + 3 O2
From the balanced equation, we can see that for every 2 moles of KClO3, we get 3 moles of O2. We can use the volume ratio between gases at STP (22.4 L = 1 mole of any gas) to convert the desired volume of O2 into moles:
5.00 L O2 x (1 mole O2 / 22.4 L O2) = 0.223 mols O2
Now, using the mole ratio from the balanced equation, we can calculate the moles of KClO3 needed to produce the desired amount of O2:
0.223 mol O2 x (2 mol KClO3 / 3 mol O2) = 0.1487 mol KClO3
To convert this into grams of KClO3, we use the molar mass of KClO3 which is approximately 122.55 g/mol:
0.1487 mol KClO3 x 122.55 g/mol = 18.20 g KClO3
Therefore, approximately 18.20 grams of KClO3 are needed to produce 5.00 L of oxygen gas at STP.
Now, let's calculate the percent yield. In this case, we know that the reaction produces 4.80 L of O2.
Using the same stoichiometry as before, we can calculate the theoretical yield of O2:
4.80 L O2 x (1 mole O2 / 22.4 L O2) = 0.2143 mol O2
Now, since the moles of KClO3 and O2 have a 2:3 ratio, we can calculate the theoretical moles of KClO3:
0.2143 mol O2 x (2 mol KClO3 / 3 mol O2) = 0.1429 mol KClO3
Comparing this to the actual yield, we can calculate the percent yield:
Percent yield = (Actual yield / Theoretical yield) x 100
= (0.2143 mol O2 / 0.1429 mol KClO3) x 100
= 150%
Therefore, the percent yield is 150%.