Pri reakciji koja se odigrava izmedju kalijum-permanganata i sumporvodonika u prisustvu sumporne kiseline nastaje: mangan(II)-sulfat, sumpor, kalijum-sulfat i voda.Koliko ce se dobiti miligrama sumpora uvodjenjem 112mL sumporvodonika(pri n.u) u rasvor kalijum-permanganata?
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Oksido-redukcija
2KMnO4+5H2S+3H2SO4-------->2MnSO4+5S+K2SO4+8H2O
S2-(-2e-)------------->S/5
Mn7+(-5e-)------------>Mn2+/2
1mol------------------22.4dm3
xmol-------------------0.112dm3
x=0.112*1/22.4=0.005 mol H2S
n(S):n(H2S)=5:5
n(S)=n(H2S)*5/5=0.005*5/5=0.005 mol
m(S)=n(S)*Mr=0.005*32=0.16 g=160 mg
2KMnO4+5H2S+3H2SO4-->2MnSO4+5
2KMnO4+5H2S+3H2SO4-->2MnSO4+5S+K2SO4+8H2O
...................112 ml.....................................x mg
...................22.4 ml....................................32mg
x=160 mg S
Pozdrav! :)
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