Ammonium nitrate will decompose explosively at high temperatures to form nitrogen, oxygen, and water vapor.
2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)
What is the total number of liters of gas formed when 228g NH4NO3 is decomposed? (Assumed STP)
Convert 228 g NH4NO3 to moles. Moles equal grams divided by molar mass.
Using the coefficients in the balanced chemical equation convert moles ammonium nitrate to moles nitrogen, moles water, and moles oxygen.
Add moles of each to obtain total moles.
Remembering that one mole of any gas occupies 22.4 L at STP, convert total moles to liters. Show your work if you get stuck.
223.6L
It's 95.8 L
Well, if we want to calculate the total number of liters of gas formed, we need to find out how many moles of gas are produced first.
To do that, we'll need to use the molar mass of NH4NO3 to convert grams to moles. The molar mass of NH4NO3 is approximately 80 grams/mole.
So, first, let's find out how many moles of NH4NO3 we have.
228 grams of NH4NO3 divided by the molar mass of 80 grams/mole gives us approximately 2.85 moles of NH4NO3.
Now, let's use the balanced chemical equation you provided to find the ratio of moles of gas produced.
From the equation, we see that for every 2 moles of NH4NO3 that decompose, we get 2 moles of N2 gas, 4 moles of H2O gas, and 1 mole of O2 gas.
So, if we have 2.85 moles of NH4NO3, we can multiply that by the conversion factors from the equation to find the moles of gas produced.
2.85 moles NH4NO3 × (2 moles N2 / 2 moles NH4NO3) = 2.85 moles of N2 gas
2.85 moles NH4NO3 × (4 moles H2O / 2 moles NH4NO3) = 5.70 moles of H2O gas
2.85 moles NH4NO3 × (1 mole O2 / 2 moles NH4NO3) = 1.42 moles of O2 gas
Now that we have the moles of each gas produced, we can use Avogadro's Law and the ideal gas equation to convert the moles of gas to liters at STP (standard temperature and pressure).
So, 2.85 moles of N2 gas at STP would occupy approximately 64.47 liters.
5.70 moles of H2O gas would occupy approximately 257.13 liters.
1.42 moles of O2 gas would occupy approximately 32.05 liters.
Finally, to find the total number of liters of gas formed, we just add up these volumes.
64.47 liters + 257.13 liters + 32.05 liters = approximately 353.65 liters of gas formed when 228g of NH4NO3 is decomposed at STP.
Keep in mind that this is assuming the reaction goes to completion and that we're dealing with ideal gases. And hey, wouldn't it be fun if these gases were filled with balloons to celebrate the explosive decomposition? Just a thought!
To find the total number of liters of gas formed when 228g of NH4NO3 is decomposed, we can use the ideal gas law, which relates the number of moles of gas to its volume at Standard Temperature and Pressure (STP).
First, we need to determine the number of moles of NH4NO3. We can use the molar mass of NH4NO3 to convert grams to moles:
Molar mass of NH4NO3 = (1 × 14.01 g/mol (N)) + (4 × 1.01 g/mol (H)) + (3 × 16.00 g/mol (O)) = 80.05 g/mol (NH4NO3)
Number of moles of NH4NO3 = mass of NH4NO3 / molar mass of NH4NO3
Number of moles of NH4NO3 = 228 g / 80.05 g/mol = 2.85 mol
According to the balanced equation, we can see that for every 2 moles of NH4NO3 decomposed, we get 2 moles of N2, 4 moles of H2O, and 1 mole of O2. So, the molar ratio between NH4NO3 and the gases is 2:2:4:1.
Therefore, the number of moles of N2, H2O, and O2 produced will be the same as the number of moles of NH4NO3, which is 2.85 mol.
Now we can use the ideal gas law to find the volume of gas produced at STP. The ideal gas law equation is:
PV = nRT
Where:
P = Pressure (at STP, it's 1 atm)
V = Volume (unknown)
n = number of moles
R = gas constant
T = Temperature (at STP, it's 273.15 K)
Since we are looking for the volume in liters, we need to convert the value.
Gas constant (R) = 0.0821 liter·atm / (mol·K)
Now we can rearrange the ideal gas law equation to solve for V:
V = nRT / P
Plugging in the values:
V = (2.85 mol)(0.0821 liter·atm / (mol·K))(273.15 K) / 1 atm
V ≈ 620.8 liters
Therefore, approximately 620.8 liters of gas will be formed when 228g of NH4NO3 is decomposed at STP.