Cl2 + 2HBr → 2HCl + Br2
Assuming STP, what is the volume of bromine gas produced if 82.60 liters of Cl2 react with HBr?
the same number of mols of Br2 as of Cl2
Therefore the same number of liters at STP, 82.60 liters
(about 22.4 mols/liter at STP)
sorry, mind failing, 22.4 liters /mol
luckily irrelevant for this problem
To calculate the volume of bromine gas produced, we need to use the molar ratio of Cl2 to Br2 from the balanced chemical equation.
From the equation: Cl2 + 2HBr → 2HCl + Br2
We can see that for every 1 mole of Cl2, 1 mole of Br2 is produced. Thus, the ratio is 1:1.
First, we need to calculate the number of moles of Cl2 in 82.60 liters:
Using the ideal gas law equation: PV = nRT
Assuming STP (Standard Temperature and Pressure):
P = 1 atm
T = 273.15 K
R = 0.0821 L.atm/mol.K
Using the provided volume:
V = 82.60 L
Plugging in the values into the ideal gas law equation:
(1 atm)(82.60 L) = n(0.0821 L.atm/mol.K)(273.15 K)
Solving for n:
n = (1 atm * 82.60 L) / (0.0821 L.atm/mol.K * 273.15 K)
n = 3.31 moles of Cl2
Since the molar ratio of Cl2 to Br2 is 1:1, we have 3.31 moles of Br2 produced.
Now, to calculate the volume of bromine gas produced, we can use the ideal gas law again, using the number of moles of Br2:
P = 1 atm
T = 273.15 K
R = 0.0821 L.atm/mol.K
n = 3.31 moles
Using the ideal gas law equation:
PV = nRT
Solving for V (volume):
V = (nRT) / P
V = (3.31 mol * 0.0821 L.atm/mol.K * 273.15 K) / 1 atm
Calculating:
V = 76.21 Liters
Therefore, assuming STP, the volume of bromine gas produced when 82.60 liters of Cl2 react with HBr is 76.21 liters.
To determine the volume of bromine gas produced in the reaction, we need to use the stoichiometry of the balanced chemical equation.
The balanced equation is:
Cl2 + 2HBr → 2HCl + Br2
From the balanced equation, we can see that 1 mole of Cl2 reacts with 2 moles of HBr to produce 1 mole of Br2. Additionally, we need to consider that the reaction takes place at Standard Temperature and Pressure (STP), where the temperature is 0 degrees Celsius (273.15 K) and the pressure is 1 atmosphere (atm).
Given that we have 82.60 liters of Cl2, we first need to convert this volume to moles using the Ideal Gas Law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
For STP conditions, the pressure is 1 atm and the temperature is 273.15 K, so we have:
(1 atm)(82.60 L) = n(0.0821 L·atm/mol·K)(273.15 K)
Simplifying this equation, we find:
n = (1 atm)(82.60 L) / (0.0821 L·atm/mol·K)(273.15 K)
n = 3.234 mol
Since 1 mole of Cl2 produces 1 mole of Br2, we know that 3.234 moles of Cl2 react to produce 3.234 moles of Br2.
Now, we can use the Ideal Gas Law again to calculate the volume of bromine gas produced. The number of moles of gas and the temperature remain the same as in the previous calculation, but the pressure is not specified. Assuming standard atmospheric pressure of 1 atm, we have:
PV = nRT
(1 atm)(V) = (3.234 mol)(0.0821 L·atm/mol·K)(273.15 K)
Simplifying this equation, we find:
V = (3.234 mol)(0.0821 L·atm/mol·K)(273.15 K) / 1 atm
V ≈ 70.49 L
Therefore, approximately 70.49 liters of bromine gas would be produced in this reaction under STP conditions.