To determine the grams of oxygen gas (O₂) produced in a reaction, we need to apply the Ideal Gas Law, which is expressed as PV = nRT. Here’s how to solve it step by step.
Step 1: Identify the Variables
- P = Pressure (in atm)
- V = Volume (in liters)
- n = Number of moles of gas
- R = Ideal Gas Constant (0.0821 L·atm/(K·mol))
- T = Temperature (in Kelvin)
Step 2: Convert Temperature
The temperature of tap water is given as 23 degrees Celsius. To convert this to Kelvin, we use the formula:
T(K) = T(°C) + 273.15
T(K) = 23 + 273.15 = 296.15 K
Step 3: Rearrange Ideal Gas Law to Solve for n
We can rearrange the formula to solve for n (the number of moles):
n = PV / RT
Step 4: Insert the Values
Now we need to know the pressure (P) and volume (V) of the gas produced in the reaction. Let’s say the reaction is conducted at standard atmospheric pressure (1 atm) and produces 1 L of gas:
P = 1 atm, V = 1 L
Now we substitute these values into our rearranged equation:
n = (1 atm)(1 L) / (0.0821 L·atm/(K·mol) * 296.15 K)
Calculating this gives:
n ≈ 0.0408 moles of O₂
Step 5: Convert Moles to Grams
To convert moles of O₂ to grams, we use the molar mass of O₂. The molar mass of O₂ (Oxygen) is approximately 32 g/mol:
Grams = n × Molar Mass
Grams = 0.0408 moles × 32 g/mol ≈ 1.3064 grams
Conclusion
In summary, if the reaction produces 1 L of oxygen gas at 1 atm pressure and 23 degrees Celsius, approximately 1.31 grams of O₂ will be produced.