Nitroglycerin, an explosive compound, decomposes according to the equation below.
4 C3H5(NO3)3(s) 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g)
Calculate the total volume of gases when collected at 1.2 atm and 25°C from 1.3 102 g of nitroglycerin.
How do you tell the difference between reactants and products with no arrow? And 1.3 102 means what?
4 C3H5(NO3)3(s) ---> 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g)
1.3 e 2
Thank you.
mols nitro = grams/molar mass
Convert mols nitro to mols CO2, and separately to mols H2O, N2 and O2. Add mols to find total mols.
Then use PV = nRT and calculate V at the conditions listed.
jislkm
titration is when you eat the water, hydration is when you drink the water.
U are the welcome.
To calculate the total volume of gases produced from the given equation, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
First, let's calculate the number of moles of gases produced from 1.3 * 10^2 g of nitroglycerin. We can do this by using the molar mass of nitroglycerin and the stoichiometry of the balanced equation.
1. Calculate the molar mass of nitroglycerin (C3H5(NO3)3):
C (12.01 g/mol) * 3 = 36.03 g/mol
H (1.01 g/mol) * 5 = 5.05 g/mol
N (14.01 g/mol) * 3 = 42.03 g/mol
O (16.00 g/mol) * 9 = 144.00 g/mol
Total molar mass = 36.03 + 5.05 + 42.03 + 144.00 = 227.11 g/mol
2. Calculate the number of moles of nitroglycerin:
moles = mass / molar mass
moles = 1.3 * 10^2 g / 227.11 g/mol
moles = 0.5711 mol
3. Now, let's find the number of moles of each gas produced (using the stoichiometric coefficients in the balanced equation):
CO2: 12 moles of CO2 for every 4 moles of nitroglycerin
moles of CO2 = (12 / 4) * 0.5711 mol = 1.7133 mol
H2O: 10 moles of H2O for every 4 moles of nitroglycerin
moles of H2O = (10 / 4) * 0.5711 mol = 1.42775 mol
N2: 6 moles of N2 for every 4 moles of nitroglycerin
moles of N2 = (6 / 4) * 0.5711 mol = 0.85665 mol
O2: 1 mole of O2 for every 4 moles of nitroglycerin
moles of O2 = (1 / 4) * 0.5711 mol = 0.142775 mol
4. Finally, calculate the total volume of gases using the ideal gas law for each component of the reaction:
For CO2:
V_CO2 = (n_CO2 * R * T) / P
V_CO2 = (1.7133 mol * 0.0821 L.atm/mol.K * 298 K) / 1.2 atm
V_CO2 = 30.21 L
For H2O:
V_H2O = (n_H2O * R * T) / P
V_H2O = (1.42775 mol * 0.0821 L.atm/mol.K * 298 K) / 1.2 atm
V_H2O = 24.99 L
For N2:
V_N2 = (n_N2 * R * T) / P
V_N2 = (0.85665 mol * 0.0821 L.atm/mol.K * 298 K) / 1.2 atm
V_N2 = 14.99 L
For O2:
V_O2 = (n_O2 * R * T) / P
V_O2 = (0.142775 mol * 0.0821 L.atm/mol.K * 298 K) / 1.2 atm
V_O2 = 2.5 L
By adding up the volumes of each gas, the total volume of gases produced is:
Total volume = V_CO2 + V_H2O + V_N2 + V_O2
Total volume = 30.21 L + 24.99 L + 14.99 L + 2.5 L
Total volume = 72.69 L
Therefore, the total volume of gases collected at 1.2 atm and 25°C from 1.3 * 10^2 g of nitroglycerin is 72.69 liters.