What is the total volume of gases produced at 819 K and 1. atm pressure when 320 g of ammonium nitrite undergoes the following decomposition reaction?
NH4NO2 --------> N2(g) + 2H2O(g)
Well, if I had a nose, I'd say this question stinks! But lucky for me, I don't have a nose. Math, on the other hand, is more up my alley, or in this case, up my red, bulbous nose.
To find the total volume of gases produced, we need to use the ideal gas law equation, which goes like this: PV = nRT.
First things first, we need to find the number of moles (n) of the product gases.
From the balanced equation, we can see that for every 1 mole of NH4NO2, 1 mole of N2 and 2 moles of H2O are produced. So, let's calculate the number of moles of NH4NO2:
molar mass of NH4NO2 = 80 g/mol
moles of NH4NO2 = 320 g / 80 g/mol
moles of NH4NO2 = 4 moles
Now that we know the number of moles of NH4NO2, we can determine the number of moles of the product gases. Since the stoichiometry shows that 1 mole of NH4NO2 produces 1 mole of N2 and 2 moles of H2O, we get:
moles of N2 = 4 moles
moles of H2O = 2 x 4 moles
moles of H2O = 8 moles
Now we can plug these values into the ideal gas law equation to find the total volume of gases. But hey, let's not forget the other variables:
- P = 1 atm (pressure)
- V = ? (volume)
- n = 4 moles (number of moles)
- R = 0.0821 L·atm/(mol·K) (the ideal gas constant)
- T = 819 K (temperature)
Rearranging the ideal gas law equation to solve for volume (V), we get:
V = (nRT) / P
V = (4 moles x 0.0821 L·atm/(mol·K) x 819 K) / 1 atm
Crunching the numbers, we find that the total volume of gases produced is around 273.1 liters.
So there you have it! The volume of gases produced at 819 K and 1 atm pressure when 320 g of ammonium nitrite decomposes is approximately 273.1 liters. Now I'm just as deflated as a popped balloon. Or maybe I just need to put on a happy face and juggle some more calculations!
To find the total volume of gases produced at 819 K and 1 atm pressure, we will 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.
First, we need to calculate the number of moles of N2 and H2O produced from 320 g of ammonium nitrite (NH4NO2).
The molar mass of NH4NO2 = 80.052 g/mol.
So, the number of moles of NH4NO2 = 320 g / 80.052 g/mol = 3.998 mol (approximately 4 mol).
According to the balanced equation, 1 mol of NH4NO2 produces 1 mol of N2 and 2 mol of H2O.
Therefore, the number of moles of N2 = 4 mol, and the number of moles of H2O = 2 x 4 mol = 8 mol.
Now we can calculate the total volume of gases using the ideal gas law.
First, we need to convert the temperature from Celsius to Kelvin:
819 K (Kelvin) = 819 °C + 273.15 = 1092.15 K.
Next, we can calculate the total volume of gases produced:
For N2:
P1V1 = n1RT1
1 atm x V1 = 4 mol x (0.0821 L·atm/mol·K) x 1092.15 K
V1 = (4 mol x 0.0821 L·atm/mol·K x 1092.15 K) / 1 atm
V1 = 366.382 L
For H2O:
P2V2 = n2RT2
1 atm x V2 = 8 mol x (0.0821 L·atm/mol·K) x 1092.15 K
V2 = (8 mol x 0.0821 L·atm/mol·K x 1092.15 K) / 1 atm
V2 = 732.764 L
Therefore, the total volume of gases produced at 819 K and 1 atm pressure when 320 g of ammonium nitrite undergoes the given decomposition reaction is approximately 366.382 L (N2) + 732.764 L (H2O) = 1099.146 L.
To find the total volume of gases produced, we need to use the ideal gas law equation:
PV = nRT
Where:
P = Pressure of the gas (1. atm)
V = Volume of the gas (we need to find this)
n = Number of moles of the gas (we need to find this)
R = Ideal gas constant (0.0821 L.atm/mol.K)
T = Temperature in Kelvin (819 K)
First, let's calculate the number of moles of N2 gas produced:
1 mole of NH4NO2 produces 1 mole of N2 according to the balanced equation.
The molar mass of NH4NO2 = (1 * 14.01) + (4 * 1.01) + (1 * 14.01) + (2 * 16) = 80.05 g/mol.
So, the number of moles of NH4NO2 = mass / molar mass = 320 g / 80.05 g/mol = 3.998 mol ≈ 4 mol.
Therefore, the number of moles of N2 produced = 4 mol.
Now, let's calculate the number of moles of H2O gas produced:
1 mole of NH4NO2 produces 2 moles of H2O according to the balanced equation.
Therefore, the number of moles of H2O produced = 2 * (number of moles of NH4NO2) = 2 * 4 mol = 8 mol.
Since there are no other gases produced in the reaction, the total number of moles of gases produced = number of moles of N2 + number of moles of H2O = 4 mol + 8 mol = 12 mol.
Now we can use the ideal gas law equation to find the volume of the gases:
PV = nRT
V = (nRT) / P
V = (12 mol * 0.0821 L.atm/mol.K * 819 K) / 1. atm
V = 81.8196 L
Therefore, the total volume of gases produced at 819 K and 1 atm pressure when 320 g of ammonium nitrite undergoes decomposition is approximately 81.82 L.
mols NH4NO2 (is that what you meant to type?) = grams/molar mass = about 5 mols but you need to confirm that.
Therefore, you must have produced 5 mols N2 and 10 mols H2O for a total of 15 mol.
Use PV = nRT and solve for V in L.