Answer to Question #93794 in General Chemistry for adewale

Question

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.

Answer 1

a) Schematically, the reaction could be expressed with a following equation:

KMnO₄ + H₂O₂ + H₂SO₄ $\longrightarrow$ K₂SO₄ + MnSO₄ + O₂ + H₂O

The half-reactions in this red-ox system would be:

$MnO_{4}^-+8H^++5e\longrightarrow Mn^{2+}+4H_{2}O$

$H_{2}O_{2}-2e \longrightarrow O_{2}+2H^+$

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:

$2MnO_{4}^-+16H^++5H_{2}O_{2} \longrightarrow 2Mn^{2+}+8H_{2}O + 5O_{2}+10H^+$

Now, we cancel some H⁺ cations from both sides and get this:

$2MnO_{4}^-+6H^++5H_{2}O_{2} \longrightarrow 2Mn^{2+} + 8H_{2}O + 5O_{2}$

This is short ionic equation for the following molecular equation:

$2KMnO_{4}+5H_{2}O_{2}+3H_{2}SO_{4} \longrightarrow 2MnSO_{4}+K_{2}SO_{4}+5O_{2}+8H_{2}O$

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:

$n(KMnO_{4})=\frac{m(KMnO_{4})}{M(KMnO_{4})}=\frac{100g}{158g/mol}=0.633mol$

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:

$PV(O_{2})=n(O_{2})RT$

$V(O_{2})=\frac{n(O_{2})RT}{P}=\frac{1.583mol*0.082atmLK^{-1}mol^{-1}*398K}{2atm}=25.8L$

Thus, there will be some 25.8 L of oxygen generated by the mentioned reaction.