If 4.10 × 102 mol of octane combusts, what volume of carbon dioxide is produced at 22.0 °C and 0.995 atm?
To solve this problem, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
In this case, we are given the number of moles (n) of octane that combusts (4.10 × 10^2 mol), the temperature (T) in degrees Celsius (22.0 °C), and the pressure (P) in atm (0.995 atm). We need to convert the temperature to Kelvin by using the equation:
T(K) = T(°C) + 273.15
So, the temperature in Kelvin is 22.0 + 273.15 = 295.15 K.
Now, we can rearrange the ideal gas law equation to solve for the volume (V):
V = (nRT) / P
Substituting the values we have:
V = (4.10 × 10^2 mol) × (0.0821 L·atm/mol·K) × (295.15 K) / (0.995 atm) = 3798.78 L
Therefore, the volume of carbon dioxide produced is 3798.78 liters at 22.0 °C and 0.995 atm when 4.10 × 10^2 mol of octane combusts.
To find the volume of carbon dioxide (CO2) produced, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure in atm
V = volume in liters
n = moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 22.0 + 273.15
T(K) = 295.15 K
Given:
n = 4.10 × 10^2 mol
T = 295.15 K
P = 0.995 atm
Now let's solve for V:
V = (nRT) / P
V = (4.10 × 10^2 mol) × (0.0821 L·atm/mol·K) × (295.15 K) / 0.995 atm
Calculating this expression gives us the volume.
2C8H18 + 25O2 ==>16CO2 + 18H2O
Convert 410 mols C8H18 to mols CO2.
410 x (16 mols CO2/2 mols C8H18) = n
Substitute n into PV = nRT at the conditions listed and solve for volume in liters.