When nitroglycerin (C3H5N3O9) explodes, it decomposes into the following gases: CO2, N2, NO, and H2O. If 239g of nitroglycerin explodes, what volume will the mixture of gaseous produces occupy at 1.00 atm pressure and 2678oC?
a. Write a balanced equation for the reaction.
b. Use stoichiometry to find the moles of ALL the gas products combined.
c. Use the ideal gas law to solve the volume of the products of the explosion.
Is A correct?
A. 4C3H5N3O9>12CO2+5N2+2NO+10H2O
Not sure about the other yet. I am trying to figure it out , so I can explain it to my daughter. Is A correct?
I didn't check the oxidation numbers and electron change but the coefficients appear to be ok.
To continue, convert 239 g nitro into mols. mol = grams/molar mass
Use the coefficients in the balanced equation to convert mols nitro to mols N2, mols H2O, mols CO2, mols NO. That's part b.
Add the mols together to find total mols, then use PV = nRT to find V. This is part C. Don't forget that P is in atmospheres and T must be in kelvin. which is 273 + C
Yes, you are correct. The balanced equation for the decomposition of nitroglycerin is:
4C3H5N3O9 -> 12CO2 + 5N2 + 2NO + 10H2O
In this equation, 4 molecules of nitroglycerin decompose to produce 12 molecules of carbon dioxide (CO2), 5 molecules of nitrogen gas (N2), 2 molecules of nitric oxide gas (NO), and 10 molecules of water (H2O).
Now, let's move on to part B, where we will use stoichiometry to find the moles of all the gas products combined.
To find the moles of all the gas products, we need to determine the moles of nitroglycerin first. You are given that 239g of nitroglycerin decomposes, so we can calculate the moles of nitroglycerin using its molar mass.
The molar mass of nitroglycerin (C3H5N3O9) can be calculated as follows:
(3*12.01 g/mol) + (5*1.01 g/mol) + (3*14.01 g/mol) + (9*16.00 g/mol) = 227.12 g/mol
Now we can find the moles of nitroglycerin by dividing the given mass by the molar mass:
239 g / 227.12 g/mol = 1.053 mol of nitroglycerin
According to the balanced equation, 4 moles of nitroglycerin produce a combination of gases. Therefore, to find the moles of the gas products, we will use the mole ratio between nitroglycerin and the gases (CO2, N2, NO, H2O) from the balanced equation.
From the equation, we can see that:
4 moles of nitroglycerin produce:
12 moles of CO2
5 moles of N2
2 moles of NO
10 moles of H2O
Since we have found that there are 1.053 moles of nitroglycerin, we can use these ratios to calculate the moles of each gas product:
CO2: 1.053 moles (nitroglycerin) * (12 moles CO2 / 4 moles nitroglycerin) = 3.159 moles CO2
N2: 1.053 moles * (5 moles N2 / 4 moles) = 1.316 moles N2
NO: 1.053 moles * (2 moles NO / 4 moles) = 0.527 moles NO
H2O: 1.053 moles * (10 moles H2O / 4 moles) = 2.632 moles H2O
Now, let's move on to part C, where we will use the ideal gas law to solve for the volume of the gas products.
To solve for the volume, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (1.00 atm)
V = volume (what we need to find)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (2678oC + 273.15 = 2951.15 K)
Now, let's substitute the values into the equation:
(1.00 atm) * V = (3.159 moles CO2 + 1.316 moles N2 + 0.527 moles NO + 2.632 moles H2O) * (0.0821 L·atm/(mol·K)) * (2951.15 K)
Simplifying the equation, we can solve for V:
V = [(3.159 + 1.316 + 0.527 + 2.632) * 0.0821 * 2951.15] / 1.00
After performing the calculations, you will find the value of V, which represents the volume of the gas mixture in liters.
Yes, the balanced equation you provided is correct.
To calculate the moles of all the gas products combined, you need to use stoichiometry.
From the balanced equation, you can see that for every 4 moles of C3H5N3O9 that explode, you get 12 moles of CO2, 5 moles of N2, 2 moles of NO, and 10 moles of H2O as products.
Now, we can convert the mass of nitroglycerin into moles. The molar mass of nitroglycerin (C3H5N3O9) is calculated as follows:
C (3 × 12.01 g/mol) + H (5 × 1.01 g/mol) + N (3 × 14.01 g/mol) + O (9 × 16.00 g/mol) = 227.09 g/mol
To find the moles of nitroglycerin, we divide the mass (239 g) by the molar mass (227.09 g/mol):
239 g / 227.09 g/mol = 1.053 mol
Now, we can use the stoichiometric ratios from the balanced equation to find the moles of all the gas products:
- CO2: 12 moles of CO2 are produced for every 4 moles of C3H5N3O9.
Moles of CO2 = (12 moles CO2 / 4 moles C3H5N3O9) × 1.053 mol C3H5N3O9 = 3.158 mol CO2
- N2: 5 moles of N2 are produced for every 4 moles of C3H5N3O9.
Moles of N2 = (5 moles N2 / 4 moles C3H5N3O9) × 1.053 mol C3H5N3O9 = 1.316 mol N2
- NO: 2 moles of NO are produced for every 4 moles of C3H5N3O9.
Moles of NO = (2 moles NO / 4 moles C3H5N3O9) × 1.053 mol C3H5N3O9 = 0.526 mol NO
- H2O: 10 moles of H2O are produced for every 4 moles of C3H5N3O9.
Moles of H2O = (10 moles H2O / 4 moles C3H5N3O9) × 1.053 mol C3H5N3O9 = 2.632 mol H2O
Now that we have the moles of all the gas products, we can use the ideal gas law to find the volume.
The ideal gas law is given by: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Since the pressure and temperature are given, we can rearrange the ideal gas law to solve for volume:
V = (nRT) / P
Using the values:
n = 3.158 mol CO2 + 1.316 mol N2 + 0.526 mol NO + 2.632 mol H2O (combined moles of all gas products)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = 2678°C = (2678 + 273) K = 2951 K (convert to Kelvin)
P = 1.00 atm (given pressure)
V = ((3.158 + 1.316 + 0.526 + 2.632) mol) × (0.0821 L·atm/(mol·K)) × (2951 K) / (1.00 atm)
V ≈ 288.6 L
Therefore, the volume occupied by the gaseous products of the explosion will be approximately 288.6 liters at a pressure of 1.00 atm and a temperature of 2678°C.