If 3.0 liters of oxygen gas react with excess carbon monoxide at STP, how many liters of carbon dioxide can be produced under the same conditions?

2 CO (g) + O2 (g) ---> 2 CO2 (g)

2CO + O2 ==> 2CO2

When using gases one may take a shortcut and use L directly as if they were mols.
3.0 L O2 x (2 mols CO2/1 mol O2) = 3 x 2/1 = 6 L CO2 produced.

To determine the amount of carbon dioxide that can be produced, we need to use the stoichiometry of the balanced chemical equation. The coefficients in the balanced equation represent the mole ratio between reactants and products.

From the balanced equation, we can see that 2 moles of CO react with 1 mole of O2 to produce 2 moles of CO2.

First, let's convert the given volume of oxygen gas to moles using the ideal gas law:

PV = nRT

Since the conditions are STP (Standard Temperature and Pressure), we can use the following values:
V = 3.0 L
P = 1 atm
T = 273 K
R = 0.0821 L.atm/mol.K

n = PV / RT
n = (1 atm)(3.0 L) / (0.0821 L.atm/mol.K)(273 K)
n ≈ 0.1376 moles

Now that we have the moles of oxygen gas, we can use the stoichiometry to calculate the moles of carbon dioxide produced.

Since the mole ratio between O2 and CO2 is 1:2, the number of moles of CO2 produced will be double the number of moles of O2 used.

n_CO2 = 2 * n_O2
n_CO2 = 2 * 0.1376 moles
n_CO2 ≈ 0.2752 moles

Finally, we can convert the moles of carbon dioxide back into volume using the ideal gas law:

V = nRT / P
V = (0.2752 moles)(0.0821 L.atm/mol.K)(273 K) / (1 atm)
V ≈ 6.556 L

Therefore, approximately 6.556 liters of carbon dioxide can be produced under the same conditions.

To find the volume of carbon dioxide produced, we first need to determine the stoichiometry of the reaction.

The balanced equation tells us that 2 mol of carbon monoxide react with 1 mol of oxygen gas to form 2 mol of carbon dioxide.

Since we are given the volume of oxygen gas (3.0 L), we need to convert it to moles. To do this, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure (STP conditions = 1 atm)
V = volume (3.0 L)
n = number of moles (unknown)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (STP conditions = 273 K)

Rearranging the equation, we have:

n = PV / RT

Substituting the values:

n = (1 atm) * (3.0 L) / (0.0821 L·atm/(mol·K) * 273 K)
n = 0.121 moles of oxygen gas

According to the stoichiometry of the balanced equation, 2 moles of carbon monoxide react with 1 mole of oxygen gas to produce 2 moles of carbon dioxide.

Therefore, the amount of carbon dioxide produced will also be 0.121 moles.

To convert moles to liters, we can use the same ideal gas law equation:

V = nRT / P

Substituting the values:

V = (0.121 moles) * (0.0821 L·atm/(mol·K)) * (273 K) / (1 atm)
V = 2.778 L

So, under the given conditions, 2.778 liters of carbon dioxide can be produced.