How many liters of Cl2 has at a pressure of .950 atm and a temperature of 298K will be collected from the reaction of 25 mL of a .100 M aqueous solution of KMnO4 and an excess of HCI?
KMnO4+HCI->MnCl2+Cl2+H2)+KCl
To find the volume of Cl2, you will need to use the Ideal Gas Law equation, which states:
PV = nRT
Where:
P = pressure in atmospheres (atm)
V = volume in liters (L)
n = number of moles (mol)
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = temperature in Kelvin (K)
First, let's find the number of moles of Cl2 produced in the reaction. From the balanced equation, we can determine that the stoichiometric ratio between KMnO4 and Cl2 is 1:1. This means that for every mole of KMnO4 reacted, one mole of Cl2 is produced.
To find the moles of KMnO4, we can use the formula:
moles = molarity x volume
Given that the volume of the KMnO4 solution is 25 mL (which is equal to 0.025 L) and the molarity of the solution is 0.100 M, we can calculate the moles of KMnO4:
moles of KMnO4 = 0.100 M x 0.025 L = 0.0025 mol
Since the stoichiometric ratio between KMnO4 and Cl2 is 1:1, the moles of Cl2 produced will also be 0.0025 mol.
Now, we can use the Ideal Gas Law equation to find the volume of Cl2:
PV = nRT
Substituting the given values:
P = 0.950 atm
V = volume of Cl2 (unknown)
n = 0.0025 mol
R = 0.0821 L·atm/(mol·K)
T = 298 K
0.950 atm × V = 0.0025 mol × 0.0821 L·atm/(mol·K) × 298 K
Simplifying the equation:
0.950 V = 0.1939
Dividing both sides by 0.950:
V = 0.1939 / 0.950 ≈ 0.204 L
Therefore, the volume of Cl2 produced is approximately 0.204 liters.
mols KMnO4 = M x L = ?
Use simple stoichiometry to convert mols KMnO4 to mols Cl2, then use PV = nRT and solve for V (in L).