10 L of 0.25 molof oxygen gas at a temperature of 30 °C was converted to ozone at the same temperature and pressure. Calculate the volume of ozone (O3) in mL? (Molar mass: O =16g/mol)
3O2 → 2O3
First, we need to find the number of moles of oxygen gas we have:
n(O2) = (0.25 mol/L) x (10 L) = 2.5 mol
According to the balanced equation, 3 moles of oxygen gas produce 2 moles of ozone. Therefore, we can use a proportion to find the number of moles of ozone produced:
n(O3) = (2/3) x n(O2) = (2/3) x 2.5 mol = 1.67 mol
Now we can use the ideal gas law to find the volume of ozone:
PV = nRT
We know the pressure, temperature, and number of moles, and we can assume that the volume of ozone is essentially the same as the volume of oxygen gas that it was produced from. Therefore, we can rearrange the ideal gas law to solve for volume:
V = nRT/P
V = (1.67 mol) x (0.0821 L·atm/mol·K) x (303 K) / (1 atm)
V = 40.8 L
However, we want the answer in mL, so we need to convert:
V = 40.8 L x (1000 mL / 1 L) = 40,800 mL
Therefore, the volume of ozone produced is 40,800 mL (or 40.8 L).
To solve this problem, we need to use the molar ratio from the balanced chemical equation and the ideal gas law.
Step 1: Calculate the number of moles of oxygen gas (O2).
Given: 10 L of 0.25 mol of oxygen gas at a temperature of 30 °C.
Step 2: Use the molar ratio from the balanced chemical equation to determine the number of moles of ozone (O3) produced.
The balanced chemical equation is: 3O2 → 2O3
From this equation, we can see that for every 3 moles of O2, we get 2 moles of O3.
Since we have 0.25 moles of O2, we can calculate the number of moles of O3:
0.25 moles O2 * (2 moles O3 / 3 moles O2) = 0.1667 moles O3
Step 3: Calculate the volume of ozone (O3) in mL using the ideal gas law.
Given: Temperature of 30 °C and pressure is assumed to be constant.
The ideal gas law equation is: PV = nRT
Where:
P is the pressure (constant)
V is the volume
n is the number of moles
R is the ideal gas constant (0.0821 L*atm/mol*K)
T is the temperature in Kelvin
To convert the given temperature from degrees Celsius (°C) to Kelvin (K):
T(K) = T(°C) + 273.15
T(K) = 30 + 273.15 = 303.15 K
We can rearrange the ideal gas law equation to solve for V (volume):
V = nRT / P
Given: R = 0.0821 L*atm/mol*K, n = 0.1667 moles, and T = 303.15 K.
Let's assume the pressure is 1 atm for simplicity.
V = (0.1667 moles) * (0.0821 L*atm/mol*K) * (303.15 K) / 1 atm
V = 4.2436 L
Step 4: Convert the volume of ozone (O3) from liters (L) to milliliters (mL).
1 L = 1000 mL
Volume of ozone (O3) in mL = 4.2436 L * 1000 mL/L
Volume of ozone (O3) in mL = 4243.6 mL
Therefore, the volume of ozone (O3) produced is approximately 4243.6 mL.