What total volume (in lhter at 600 C and 1atm.) could be formed by the decomposition of 16g of NH4NO3 ?
To determine the total volume of gas produced by the decomposition of NH4NO3, we need to follow these steps:
Step 1: Write the balanced chemical equation for the decomposition of NH4NO3:
NH4NO3 → N2O + 2H2O
According to the equation, 1 mole of NH4NO3 decomposes to form 1 mole of N2O and 2 moles of H2O.
Step 2: Calculate the number of moles of NH4NO3:
To find the number of moles, we use the formula:
moles = mass / molar mass
The molar mass of NH4NO3 can be calculated using the atomic masses of each element:
Molar mass of NH4NO3 = (1 x molar mass of N) + (4 x molar mass of H) + (3 x molar mass of O)
Molar mass of N = 14.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of NH4NO3 = (1 x 14.01) + (4 x 1.01) + (3 x 16.00) = 80.05 g/mol
Now, we can calculate the number of moles:
moles of NH4NO3 = 16 g / 80.05 g/mol = 0.1998 moles (approximately)
Step 3: Calculate the number of moles of N2O produced:
According to the balanced equation, for every 1 mole of NH4NO3 decomposed, 1 mole of N2O is produced.
moles of N2O = 0.1998 moles
Step 4: Convert moles of N2O to volume at 600°C and 1 atm:
In order to find the volume, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (1 atm)
V = volume (unknown)
n = number of moles (0.1998 moles)
R = ideal gas constant (0.0821 L·atm/(K·mol))
T = temperature in Kelvin (600°C = 600 + 273 = 873 K)
Rearranging the equation to solve for V:
V = (nRT) / P
V = (0.1998 moles x 0.0821 L·atm/(K·mol) x 873 K) / 1 atm
V ≈ 14.61 liters
Therefore, approximately 14.61 liters of gas could be formed by the decomposition of 16g of NH4NO3 at 600°C and 1 atm.
write the balanced equation
ammonium nitrate>Nitrogen + hydrogen+ oxygen
balance that.
Now how many moles of ammoniumnitrate is 16grams? What does that tell you the moles of gases you have? use that for n in PV=nRT