A 375-gallon tank is filled with water containing 167grams of bromine in the form of Br- ions. How many liters of Cl2 gas at 1.00 atm and 20C will be required to oxidize all the bromide to molecular bromine?
2Br^- + Cl2 ==> Br2 + 2Cl^-
mols Br^- 167/atomic mass Br.
Convert mols Br to mols Cl2.
Use PV = nRT an solve for V in L.
25.3 L of Cl2 needed.
Thanks DrBob222!
To find out how many liters of Cl2 gas are required to oxidize the bromide to molecular bromine, we need to use the ideal gas law equation: PV = nRT.
First, we need to convert the given temperature from Celsius to Kelvin:
T = 20°C + 273.15 = 293.15 K
Next, let's calculate the number of moles of bromine ions (Br-) using the given mass and molar mass of bromine (Br):
molar mass of Br = 79.904 g/mol
moles of Br- = mass of Br- / molar mass of Br
moles of Br- = 167 g / 79.904 g/mol
Now, we can calculate the volume of Cl2 gas needed. Since we know the temperature, pressure, and number of moles of Cl2 gas, we can rearrange the ideal gas law equation to solve for volume (V):
V = nRT / P
Given:
Pressure (P) = 1.00 atm
Temperature (T) = 293.15 K
Number of moles of Cl2 (n) = number of moles of Br-
Now, we can substitute the values into the equation:
V = (moles of Br-) * (0.0821 L·atm/(mol·K)) * (293.15 K) / (1.00 atm)
Simplifying the expression:
V = (moles of Br-) * (24.465 L/mol)
Now, let's calculate the moles of Br- and substitute it into the equation:
moles of Br- = 167 g / 79.904 g/mol
V = (167 g / 79.904 g/mol) * (24.465 L/mol)
Finally, we can solve for V:
V ≈ 51.25 L
Therefore, approximately 51.25 liters of Cl2 gas at 1.00 atm and 20°C will be required to oxidize all the bromide ions to molecular bromine in the 375-gallon tank.