What mass (in grams)of KNO3 would have to be decomposed to produce 29 L of oxygen measured at STP?
Here is a worked example of a stoichiometry problem. Remember 29L oxygen at STP is 29/22.4 = ??moles.
To calculate the mass of KNO3 needed to produce 29 L of oxygen at STP, we need to use stoichiometry.
Step 1: Write the balanced equation for the decomposition of KNO3.
2 KNO3(s) → 2 KNO2(s) + O2(g)
According to the balanced equation, 2 moles of KNO3 produce 1 mole of O2.
Step 2: Convert liters of oxygen to moles using the ideal gas law at STP.
1 mole of any gas occupies 22.4 L at STP. Therefore, 29 L of oxygen is equal to:
29 L * (1 mole/22.4 L) = 1.295 moles of O2
Step 3: Use the stoichiometry from the balanced equation to convert moles of O2 to moles of KNO3.
From the balanced equation, we know that 2 moles of KNO3 produce 1 mole of O2. Therefore:
1.295 moles O2 * (2 moles KNO3/1 mole O2) = 2.59 moles KNO3
Step 4: Convert moles of KNO3 to grams using the molar mass of KNO3.
The molar mass of KNO3 is 101.1 g/mol.
2.59 moles KNO3 * (101.1 g/1 mole KNO3) = 261.849 g
So, approximately 261.849 grams of KNO3 would have to be decomposed to produce 29 L of oxygen at STP.
To find the mass of KNO3 that needs to be decomposed, we first need to calculate the number of moles of oxygen that occupy 29 L at STP (Standard Temperature and Pressure).
STEP 1: Convert volume to moles of oxygen
We can use the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. At STP, the pressure is 1 atm and the temperature is 273 K.
PV = nRT
(1 atm)(29 L) = n(0.0821 L.atm/mol.K)(273 K)
29 atm.L = n(22.414 L)
n = (29 atm.L) / (22.414 L/mol)
n ≈ 1.29 mol
Therefore, 29 L of oxygen at STP corresponds to approximately 1.29 mol of oxygen.
STEP 2: Convert moles of oxygen to moles of KNO3
From the balanced chemical equation for the decomposition of KNO3, we know that 2 moles of KNO3 produce 3 moles of oxygen.
So, to find the moles of KNO3 required to produce 1.29 mol of oxygen, we can use a ratio:
2 mol KNO3 : 3 mol O2
Let's calculate it:
(1.29 mol O2) x (2 mol KNO3 / 3 mol O2) ≈ 0.86 mol KNO3
Therefore, approximately 0.86 mol of KNO3 is required to produce 29 L of oxygen at STP.
STEP 3: Convert moles of KNO3 to grams
Finally, to find the mass of KNO3 needed, we can use its molar mass. The molar mass of KNO3 can be calculated by summing the atomic masses for each element: potassium (K), nitrogen (N), and oxygen (O).
KNO3:
K: 1 atom x 39.10 g/mol = 39.10 g/mol
N: 1 atom x 14.01 g/mol = 14.01 g/mol
O: 3 atoms x 16.00 g/mol = 48.00 g/mol
Total molar mass of KNO3: 39.10 g/mol + 14.01 g/mol + 48.00 g/mol = 101.11 g/mol
Now, let's calculate the mass of KNO3 needed:
(0.86 mol KNO3) x (101.11 g KNO3 / 1 mol KNO3) ≈ 87.02 g
Therefore, approximately 87.02 grams of KNO3 would need to be decomposed to produce 29 L of oxygen at STP.