Consider the following chemical equation.
Cl2(g) + O2(g) → Cl2O3(g)
Assuming all gases are at the same temperature and pressure, how many milliliters of oxygen gas must react to give 3.50 L of Cl2O3?
same process.
To calculate the volume of oxygen gas required to produce 3.50 L of Cl2O3, we need to use the stoichiometry of the reaction and the ideal gas law.
First, let's examine the balanced chemical equation:
Cl2(g) + O2(g) → Cl2O3(g)
From the equation, we can see that the stoichiometric ratio between Cl2 and O2 is 1:1. This means that for every 1 mole of Cl2, we need 1 mole of O2.
To find the number of moles of Cl2O3, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (assumed to be constant in this case)
V = volume of Cl2O3 (3.50 L)
n = number of moles of Cl2O3
R = ideal gas constant
T = temperature (assumed to be constant too)
The ideal gas law equation can also be rearranged to solve for n:
n = PV / RT
Once we know the number of moles of Cl2O3, we can determine the number of moles of O2 required by comparing the stoichiometric ratio from the balanced chemical equation.
Now, let's walk through the steps to calculate the volume of oxygen gas required.
1. Convert the volume of Cl2O3 from liters to milliliters:
3.50 L * 1000 mL/L = 3500 mL
2. Calculate the number of moles of Cl2O3 using the ideal gas law equation:
n = (P * V) / (R * T)
Assuming we have the necessary information for the ideal gas law (pressure and temperature), plug in the values and calculate the number of moles of Cl2O3.
3. Use the stoichiometric ratio to find the number of moles of O2:
Since Cl2 and O2 have a 1:1 mole ratio, the number of moles of O2 is the same as the number of moles of Cl2O3.
4. Finally, convert the number of moles of O2 to volume in milliliters:
Use the ideal gas law equation to convert the number of moles of O2 to volume by rearranging the equation:
V = (n * R * T) / P
Plug in the calculated number of moles of O2, the ideal gas constant (R), and the appropriate pressure and temperature values to find the volume in milliliters.
Following these steps will allow you to determine the volume of oxygen gas required to produce 3.50 L of Cl2O3.