Nitrogen can also be produced from sodium metal and potassium nitrate by the reaction:
Na(s) +KNO3 (s) --> K2O(s) +Na2O(s) + N2(g)
If we generate oxygen by the thermal decomposition of potassium chlorate:
2KClO3(s) --> 2KCl(s) +3O2(g)
How many grams of potassium nitrate and how many grams of potassium chlorate would be needed to make 200.0L of gas containing 160 L N2 and 40.0 L O2 at 1.00 atm and 290 K?
To determine the number of grams of potassium nitrate (KNO3) and potassium chlorate (KClO3) needed to produce the given quantities of gas, we can use the stoichiometry of the given reactions.
Let's start by calculating the number of moles of N2 and O2 that we need to produce.
Using the ideal gas law, we can convert the given volumes of gas (160 L N2 and 40.0 L O2) to moles at the given conditions (1.00 atm and 290 K).
1. Convert N2 volume to moles:
PV = nRT
n = PV / RT
Using the values:
P = 1.00 atm
V = 160 L
R = 0.0821 L·atm/(mol·K)
T = 290 K
n(N2) = (1.00 atm) * (160 L) / (0.0821 L·atm/(mol·K)) * (290 K) = 6.491 moles N2
2. Convert O2 volume to moles:
Similarly, we can calculate the moles of O2 using the given volume and conditions.
Using the values:
P = 1.00 atm
V = 40.0 L
R = 0.0821 L·atm/(mol·K)
T = 290 K
n(O2) = (1.00 atm) * (40.0 L) / (0.0821 L·atm/(mol·K)) * (290 K) = 1.623 moles O2
Now, let's calculate the number of moles of KNO3 and KClO3 required using the stoichiometry of the reactions.
1. For the reaction: Na(s) + KNO3(s) → K2O(s) + Na2O(s) + N2(g)
We can see that 1 mole of KNO3 produces 1 mole of N2.
Therefore, the moles of KNO3 required = moles of N2 = 6.491 moles
2. For the reaction: 2KClO3(s) → 2KCl(s) + 3O2(g)
We can see that 2 moles of KClO3 produce 3 moles of O2.
Therefore, the moles of KClO3 required = (3 moles O2 / 2 moles KClO3) * (1.623 moles O2) = 2.435 moles
Now, we can calculate the mass of KNO3 and KClO3 required using their respective molar masses.
The molar mass of KNO3 = 101.1 g/mol
The molar mass of KClO3 = 122.5 g/mol
1. Mass of KNO3 required = moles of KNO3 * molar mass of KNO3 = 6.491 moles * 101.1 g/mol = 656.63 g
2. Mass of KClO3 required = moles of KClO3 * molar mass of KClO3 = 2.435 moles * 122.5 g/mol = 298.43 g
Therefore, to produce 200.0 L of gas containing 160 L N2 and 40.0 L O2 at the given conditions, you would need 656.63 grams of potassium nitrate (KNO3) and 298.43 grams of potassium chlorate (KClO3).
To determine the grams of potassium nitrate (KNO3) and potassium chlorate (KClO3) needed to make 200.0L of gas containing 160 L N2 and 40.0 L O2 at 1.00 atm and 290 K, you need to do the following steps:
Step 1: Calculate the moles of nitrogen gas (N2):
Since 1 mole of any gas at STP (standard temperature and pressure) occupies 22.4 L, we can calculate the number of moles of N2 gas using the ratio:
160 L N2 × (1 mol N2 / 22.4 L) = 7.14 mol N2
Step 2: Calculate the moles of oxygen gas (O2):
Similarly, we can calculate the moles of O2 gas:
40.0 L O2 × (1 mol O2 / 22.4 L) = 1.79 mol O2
Step 3: Determine the molar ratio between N2 and KNO3:
From the balanced equation for the reaction between Na and KNO3, we know that 1 mole of N2 is produced for every 1 mole of KNO3 consumed. Therefore, the moles of KNO3 required is also 7.14 mol.
Step 4: Calculate the mass of KNO3:
The molar mass of KNO3 is:
K = 39.10 g/mol
N = 14.01 g/mol
O = 16.00 g/mol
K = 39.10 g/mol × 2 = 78.20 g/mol
N = 14.01 g/mol × 2 = 28.02 g/mol
O = 16.00 g/mol × 3 = 48.00 g/mol
Total molar mass of KNO3 = 78.20 g/mol + 28.02 g/mol + 48.00 g/mol = 154.22 g/mol
Mass of KNO3 = 154.22 g/mol × 7.14 mol = 1101.15 g
Therefore, you would need approximately 1101.15 grams of potassium nitrate.
Step 5: Determine the molar ratio between O2 and KClO3:
From the balanced equation for the thermal decomposition of KClO3, we know that 3 moles of O2 are produced for every 2 moles of KClO3 consumed. Therefore, the moles of KClO3 required can be calculated as follows:
Moles of O2 = 1.79 mol
Molar ratio O2/KClO3 = 3 mol/2 mol
Therefore, the moles of KClO3 required = (1.79 mol × 2 mol)/3 mol = 1.19 mol KClO3
Step 6: Calculate the mass of KClO3:
The molar mass of KClO3 is:
K = 39.10 g/mol
Cl = 35.45 g/mol
O = 16.00 g/mol × 3 = 48.00 g/mol
Total molar mass of KClO3 = 39.10 g/mol + 35.45 g/mol + 48.00 g/mol = 122.55 g/mol
Mass of KClO3 = 122.55 g/mol × 1.19 mol = 145.96 g
Therefore, you would need approximately 145.96 grams of potassium chlorate.
To summarize, you would need approximately 1101.15 grams of potassium nitrate (KNO3) and 145.96 grams of potassium chlorate (KClO3) to produce 200.0L of gas containing 160 L N2 and 40.0 L O2 at 1.00 atm and 290 K.