Given the following reaction for photosynthesis 6C02+ 6H2O→C6H12O6 +6O2 how many liters of O2 can be produced from 6.5 mol of O2 standard temperature and pressure?
To determine how many liters of O2 can be produced from 6.5 mol of O2 at standard temperature and pressure (STP), we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (standard pressure at STP is 1 atm)
V = volume
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (273.15 K)
First, we need to determine the volume of 6.5 mol of O2 using the ideal gas law equation. Since the equation is balanced with a 1:6 ratio between O2 and C6H12O6, the number of moles of O2 is equal to the number of moles of C6H12O6.
n(O2) = 6.5 mol
Using the ideal gas law equation, we can rearrange it to solve for volume (V):
V = nRT / P
Substituting the values:
V(O2) = (6.5 mol) * (0.0821 L·atm/mol·K) * (273.15 K) / (1 atm)
Simplifying the expression:
V(O2) = 146.7565 L
Therefore, 6.5 mol of O2 at STP will occupy a volume of approximately 146.8 liters.
To determine the number of liters of O2 produced from 6.5 mol of O2 at standard temperature and pressure (STP), we need to use the ideal gas law.
The ideal gas law equation is as follows:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
At STP, the pressure is 1 atmosphere (atm) and the temperature is 273.15 Kelvin (K).
We can rearrange the ideal gas law equation to solve for volume (V):
V = nRT / P
First, we need to determine the value of the ideal gas constant (R). The ideal gas constant is 0.0821 L·atm/mol·K.
Now, let's plug in the values into the equation:
V = (6.5 mol) * (0.0821 L·atm/mol·K) * (273.15 K) / (1 atm)
Simplifying the equation:
V = 140.2365 L
Therefore, 6.5 mol of O2 at STP will produce approximately 140.24 liters of O2.