KMn04, in a sulphuric acid media, reacts with H202 to create MnS04, 02 , H20 and 1<2504. a. Adjust the molecular reaction by the ion-electron method. b. What volume of 02 measured at 1520 mm of mercury and 125 °C is obtained from 100 g of KMnO/ R = 0,082 atm·L-IC,mor1. Atomic masses: O= 16; K = 39; Mn = 55 g·mol-1.
a) Schematically, the reaction could be expressed with a following equation:
KMnO4 + H2O2 + H2SO4 K2SO4 + MnSO4 + O2 + H2O
The half-reactions in this red-ox system would be:
The LCM of the number of electrons in the Red and Ox systems is equal to 10. Thus, the factors for the both half-reactions would be 2 and 5, respectively. Now, we multiply first half-reaction with 2, the second with 5 and, finally, we add them together:
Now, we cancel some H+ cations from both sides and get this:
This is short ionic equation for the following molecular equation:
b) The molar weight of KMnO4 is equal to M(K) + M(Mn) + 4*M(O) = 39g/mol + 55g/mol + 4*16g/mol = 158g/mol.
Now, we can find the amount in moles of potassium permanganate:
According to the previously constructed molecular red-ox equation the molar amount of the oxygen gas is equal to 5/2 of potassium permanganate molar amount. Thus, this is the right time to calculate the molar amount of KMnO4:
n(KMnO4) = 5/2*0.633 mol = 1.583 mol of oxygen O2.
Finally, the volume of O2 gas could be found using the ideal gas state equation:
Thus, there will be some 25.8 L of oxygen generated by the mentioned reaction.
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