0.1 mole of kmno4 react with excess of h2o2 then volume of o2 liberated
To determine the volume of oxygen gas (O2) liberated when 0.1 mole of KMnO4 reacts with an excess of H2O2, we need to consider the balanced chemical equation for the reaction. The balanced equation is as follows:
2 KMnO4 + 5 H2O2 → 2 MnO2 + 2 KOH + 3 H2O + 5 O2
From the equation, we can see that for every 2 moles of KMnO4 consumed, 5 moles of O2 are produced. Therefore, we can set up a proportion to find the volume of O2 gas liberated.
1 mole of KMnO4 produces 5 moles of O2
0.1 mole of KMnO4 will produce x moles of O2
To find the value of x, we multiply 0.1 mole by the mole ratio:
0.1 mole KMnO4 * (5 moles O2 / 2 moles KMnO4) = 0.25 moles O2
Now, to determine the volume of O2 gas liberated, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
Since the ideal gas law equation requires all the variables except V, we need additional information such as pressure and temperature, or other data, to determine the volume of O2 gas liberated.