Nitroglycerin is a shock-sensitive liquid that
detonates by the reaction
4C3H5(NO3)3(`) −!
6N2(g) + 10H2O(g) + 12CO2(g) + O2(g)
Calculate the total volume of product gases
at 205 kPa and 263�C from the detonation of
2.9 lb (1315.4 g) of nitroglycerin.
Convert g nitro to mols. mols = grams/molar mass.
Using the coefficients convert mols nitro to mols H2O, mols CO2, mols N2, and mols O2. Add mols to obtain total mols, then use PV = nRT to calculate volume in L.
do ita ufasfkj
To calculate the total volume of product gases from the detonation of 2.9 lb (1315.4 g) of nitroglycerin, we need to use the molar ratio and ideal gas law.
Step 1: Convert the mass of nitroglycerin to moles
Molar mass of nitroglycerin (C3H5(NO3)3) = 227.088 g/mol
Number of moles of nitroglycerin = mass / molar mass
Number of moles of nitroglycerin = 1315.4 g / 227.088 g/mol
Step 2: Calculate the moles of product gases using the stoichiometric ratio
From the balanced chemical equation provided, the stoichiometric ratio of nitroglycerin to product gases is 4:10:12:1. This means that for every 4 moles of nitroglycerin, we get 10 moles of N2, 12 moles of CO2, and 1 mole of O2.
Number of moles of N2 = (4 moles of nitroglycerin) x (10 moles of N2 / 4 moles of nitroglycerin)
Number of moles of CO2 = (4 moles of nitroglycerin) x (12 moles of CO2 / 4 moles of nitroglycerin)
Number of moles of O2 = (4 moles of nitroglycerin) x (1 mole of O2 / 4 moles of nitroglycerin)
Step 3: Convert the moles of product gases to volume using the ideal gas law
The ideal gas law equation is PV = nRT, where:
P = pressure in atm
V = volume in L
n = number of moles
R = gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin
Given:
Pressure (P) = 205 kPa = 2.02 atm (convert from kPa to atm)
Temperature (T) = 263 degrees C = 536.15 K (convert from degrees Celsius to Kelvin)
Volume of N2 = (number of moles of N2) x (R constant) x (temperature in K) / (pressure in atm)
Volume of CO2 = (number of moles of CO2) x (R constant) x (temperature in K) / (pressure in atm)
Volume of O2 = (number of moles of O2) x (R constant) x (temperature in K) / (pressure in atm)
Step 4: Calculate the total volume of product gases
Total volume of product gases = Volume of N2 + Volume of CO2 + Volume of O2
Now, let's calculate these values:
Step 1:
Number of moles of nitroglycerin = 1315.4 g / 227.088 g/mol = 5.79 moles
Step 2:
Number of moles of N2 = (5.79 moles) x (10 moles of N2 / 4 moles of nitroglycerin) = 14.5 moles
Number of moles of CO2 = (5.79 moles) x (12 moles of CO2 / 4 moles of nitroglycerin) = 17.4 moles
Number of moles of O2 = (5.79 moles) x (1 mole of O2 / 4 moles of nitroglycerin) = 1.45 moles
Step 3:
Volume of N2 = (14.5 moles) x (0.0821 L·atm/(mol·K)) x (536.15 K) / (2.02 atm) = 612 L
Volume of CO2 = (17.4 moles) x (0.0821 L·atm/(mol·K)) x (536.15 K) / (2.02 atm) = 735 L
Volume of O2 = (1.45 moles) x (0.0821 L·atm/(mol·K)) x (536.15 K) / (2.02 atm) = 61 L
Step 4:
Total volume of product gases = 612 L + 735 L + 61 L = 1408 L
Therefore, the total volume of product gases from the detonation of 2.9 lb (1315.4 g) of nitroglycerin at 205 kPa and 263°C is 1408 L.
To calculate the total volume of product gases from the detonation of nitroglycerin, we need to use the ideal gas law. The ideal gas law is given by the equation:
PV = nRT
Where:
P = pressure in pascals
V = volume in cubic meters
n = number of moles of gas
R = ideal gas constant (8.314 J/(K*mol))
T = temperature in Kelvin
First, let's convert the given mass of nitroglycerin from pounds to grams:
2.9 lb * (453.59 g/lb) = 1315.4 g
Now, we need to determine the number of moles of nitroglycerin. To do this, we divide the mass of nitroglycerin by its molar mass. The molar mass of nitroglycerin (C3H5(NO3)3) can be calculated by summing up the atomic masses of its constituent elements:
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of N = 14.01 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of nitroglycerin = (3 * 12.01 g/mol) + (5 * 1.01 g/mol) + (3 * (14.01 + 3 * 16.00) g/mol)
= 227.04 g/mol
Number of moles of nitroglycerin = mass of nitroglycerin / molar mass of nitroglycerin
= 1315.4 g / 227.04 g/mol
= 5.789 mol
Now that we know the number of moles of nitroglycerin, we can calculate the number of moles of each product gas produced in the reaction. From the balanced equation, we see that for every 4 moles of nitroglycerin detonated, we get:
6 moles of N2
10 moles of H2O
12 moles of CO2
1 mole of O2
Using these ratios, we can calculate the number of moles of each product gas:
Number of moles of N2 = (6 mol N2 / 4 mol nitroglycerin) * 5.789 mol nitroglycerin
= 8.6845 mol
Number of moles of H2O = (10 mol H2O / 4 mol nitroglycerin) * 5.789 mol nitroglycerin
= 14.4725 mol
Number of moles of CO2 = (12 mol CO2 / 4 mol nitroglycerin) * 5.789 mol nitroglycerin
= 17.367 mol
Number of moles of O2 = (1 mol O2 / 4 mol nitroglycerin) * 5.789 mol nitroglycerin
= 1.44725 mol
Next, we need to convert the given temperature from degrees Celsius to Kelvin:
Temperature in Kelvin = 263°C + 273.15
= 536.15 K
Now we can calculate the volume of each gas using the ideal gas law. Rearranging the ideal gas law equation, we get:
V = (n * R * T) / P
For N2:
V(N2) = (8.6845 mol * 8.314 J/(K*mol) * 536.15 K) / (205000 Pa)
= 0.0841 m^3
For H2O:
V(H2O) = (14.4725 mol * 8.314 J/(K*mol) * 536.15 K) / (205000 Pa)
= 0.1395 m^3
For CO2:
V(CO2) = (17.367 mol * 8.314 J/(K*mol) * 536.15 K) / (205000 Pa)
= 0.1671 m^3
For O2:
V(O2) = (1.44725 mol * 8.314 J/(K*mol) * 536.15 K) / (205000 Pa)
= 0.0557 m^3
Finally, to find the total volume of product gases, we add up the individual volumes:
Total volume = V(N2) + V(H2O) + V(CO2) + V(O2)
= 0.0841 m^3 + 0.1395 m^3 + 0.1671 m^3 + 0.0557 m^3
= 0.4464 m^3
Therefore, the total volume of product gases from the detonation of 2.9 lb (1315.4 g) of nitroglycerin at 205 kPa and 263°C is approximately 0.4464 cubic meters.