Calculate the maximum number of mililiters of CO2, at 770 torr and 28 degrees celsius, that could be produced by reacting 500mL of CO2, at 760 torr and 15 deg Celsius, with 500 mL of O2 at 770 torr and 0 deg Celsius.

To calculate the maximum amount of CO2 produced in this reaction, we can use the ideal gas law and stoichiometry.

Step 1: Convert the temperatures from Celsius to Kelvin.
The temperature of CO2 is 28 degrees Celsius, so we need to add 273 to convert it to Kelvin: 28 + 273 = 301 K.
The temperature of O2 is 0 degrees Celsius, so we need to add 273 to convert it to Kelvin: 0 + 273 = 273 K.

Step 2: Convert the pressure of CO2 and O2 to atm.
The pressure of CO2 is given as 770 torr. Since 1 atm is equal to 760 torr, we can convert the pressure to atm by dividing 770 torr by 760 torr per atm: 770 torr / 760 torr/atm = 1.013 atm.
The pressure of O2 is given as 770 torr, which is already in torr. We don't need to convert it.

Step 3: Apply the ideal gas law to find the number of moles of CO2 and O2.
The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.

For CO2:
P = 1.013 atm (from step 2)
V = 500 mL = 0.5 L (given)
T = 301 K (from step 1)
R = 0.0821 L·atm/(mol·K) (constant)

Using the ideal gas law, we can solve for n (number of moles):
n = PV / RT = (1.013 atm) x (0.5 L) / (0.0821 L·atm/(mol·K) x 301 K) = 0.031 mol

For O2:
P = 770 torr = 1.013 atm (from step 2)
V = 500 mL = 0.5 L (given)
T = 273 K (from step 1)
R = 0.0821 L·atm/(mol·K) (constant)

Using the ideal gas law, we can solve for n (number of moles):
n = PV / RT = (1.013 atm) x (0.5 L) / (0.0821 L·atm/(mol·K) x 273 K) = 0.032 mol

Step 4: Determine the stoichiometry of the reaction.
From the balanced equation, we can see that the reaction is:
CO2 + O2 → CO2

The stoichiometry of the reaction is 1:1, meaning 1 mole of CO2 reacts with 1 mole of O2 to produce 1 mole of CO2.

Step 5: Determine the limiting reactant.
To determine the limiting reactant, we compare the moles of CO2 and O2 calculated in step 3. Since the amount of CO2 is 0.031 mol and the amount of O2 is 0.032 mol, O2 is the limiting reactant because it is present in a smaller amount.

Step 6: Calculate the maximum amount of CO2 that can be produced.
Since O2 is the limiting reactant, all of the O2 will be consumed in the reaction. Therefore, the amount of CO2 produced will be equal to the amount of O2 used, which is 0.032 mol.

Step 7: Convert the amount of CO2 to volume in mL.
To convert moles to volume, we need to use the ideal gas law again.

For CO2:
P = 770 torr = 1.013 atm (from step 2)
V = unknown (we want to solve for this)
T = 28 degrees Celsius + 273 = 301 K (given)
n = 0.032 mol (from step 6)
R = 0.0821 L·atm/(mol·K) (constant)

Using the ideal gas law, we can solve for V (volume):
V = nRT / P = (0.032 mol) x (0.0821 L·atm/(mol·K)) x (301 K) / (1.013 atm) = 0.765 L

Finally, we can convert liters to milliliters by multiplying by 1000:
V = 0.765 L x 1000 mL / 1 L = 765 mL

Therefore, the maximum number of milliliters of CO2 that could be produced is 765 mL.