Calculate the volume of nitrogen that will be produced at S.T.P from the decomposition of 9.6g of ammonium dioxonitrate 3?
Did you mean NH4NO2. What you wrote is not a correct IUPAC name for NH4NO2 but I'll play the game.
NH4NO2 ==> N2 + 2H2O
mols NH4NO2 = 9.6/molar mass NH4NO2.
Then mols NH4NO2 = mols N2 produced.
Each mol N2 @ STP will occupy 22.4 L
Post your work if you get stuck.
To calculate the volume of nitrogen gas produced at standard temperature and pressure (STP) from the decomposition of ammonium dioxonitrate 3, we need to follow these steps:
Step 1: Write the balanced chemical equation for the decomposition of ammonium dioxonitrate 3. The balanced equation is:
2NH₄N₂O₄ → NH₄NO₂ + N₂ + 2H₂O
Step 2: Calculate the molar mass of ammonium dioxonitrate 3.
NH₄N₂O₄:
(1 * 14.01 g/mol) + (2 * 14.01 g/mol) + (4 * 16.00 g/mol) + (4 * 1.01 g/mol)
= 80.04 g/mol
Step 3: Calculate the number of moles of ammonium dioxonitrate 3 using its mass and molar mass:
Number of moles = Mass / Molar mass
Number of moles = 9.6 g / 80.04 g/mol
Number of moles = 0.12 mol
Step 4: Determine the stoichiometric ratio between ammonium dioxonitrate 3 and nitrogen gas (N₂) from the balanced equation. In this case, for every 2 moles of NH₄N₂O₄, 1 mole of N₂ is produced. Therefore, the number of moles of N₂ produced will be half of the number of moles of NH₄N₂O₄.
Number of moles of N₂ = (1/2) * 0.12 mol
Number of moles of N₂ = 0.06 mol
Step 5: Apply Avogadro's law to convert moles to volume at STP.
1 mol of any ideal gas occupies 22.4 L at STP (standard temperature and pressure).
Volume of N₂ = Number of moles of N₂ * 22.4 L/mol
Volume of N₂ = 0.06 mol * 22.4 L/mol
Volume of N₂ = 1.344 L
Therefore, the volume of nitrogen gas produced at STP from the decomposition of 9.6 g of ammonium dioxonitrate 3 is 1.344 liters.
To calculate the volume of nitrogen gas produced at STP (Standard Temperature and Pressure) from the decomposition of ammonium dioxonitrate 3 (NH4NO2), we need to use the ideal gas equation, which is:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, let's determine the number of moles of ammonium dioxonitrate 3 using its molar mass.
The molar mass of NH4NO2 can be calculated as follows:
NH4NO2:
N = 14.01 g/mol
H = 1.01 g/mol (4 hydrogen atoms)
O = 16.00 g/mol (2 oxygen atoms)
Adding all the masses together gives:
Molar mass of NH4NO2 = 14.01 g/mol + (1.01 g/mol × 4) + (16.00 g/mol × 2) = 80.05 g/mol
Now, we can calculate the number of moles of NH4NO2:
Number of moles = Mass / Molar mass
Number of moles = 9.6 g / 80.05 g/mol ≈ 0.12 mol
Since the decomposition of NH4NO2 produces 2 moles of nitrogen gas (N2) for every 1 mole of NH4NO2, we can determine the number of moles of nitrogen gas:
Number of moles of N2 = 2 × number of moles of NH4NO2 = 2 × 0.12 mol = 0.24 mol
Now, let's convert the number of moles of nitrogen gas into volume at STP.
STP is defined as 1 atm of pressure and 273.15 K (0 °C) of temperature.
Using the ideal gas equation, we can rearrange it to solve for V:
V = (nRT) / P
V = (0.24 mol × 0.0821 L·atm/(mol·K) × 273.15 K) / 1 atm
V ≈ 5.97 L
Therefore, approximately 5.97 liters of nitrogen gas will be produced at STP from the decomposition of 9.6g of ammonium dioxonitrate 3.