Question #93794

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.


1
Expert's answer
2019-09-05T03:33:55-0400

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:

MnO4+8H++5eMn2++4H2OMnO_{4}^-+8H^++5e\longrightarrow Mn^{2+}+4H_{2}O

H2O22eO2+2H+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:

2MnO4+16H++5H2O22Mn2++8H2O+5O2+10H+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:

2MnO4+6H++5H2O22Mn2++8H2O+5O22MnO_{4}^-+6H^++5H_{2}O_{2} \longrightarrow 2Mn^{2+} + 8H_{2}O + 5O_{2}

This is short ionic equation for the following molecular equation:

2KMnO4+5H2O2+3H2SO42MnSO4+K2SO4+5O2+8H2O2KMnO_{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(KMnO4)=m(KMnO4)M(KMnO4)=100g158g/mol=0.633moln(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(O2)=n(O2)RTPV(O_{2})=n(O_{2})RT

V(O2)=n(O2)RTP=1.583mol0.082atmLK1mol1398K2atm=25.8LV(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.


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