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Submitted by jovaniccaelprez... on Sun, 03/30/2014 - 12:36
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koliko ce se dobiti mola sumporne kiseline uvodjenjem 17,92 ml hlora u vodeni rastvor vodonik-sulfida?

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milicaradenkovic - Sun, 03/30/2014 - 12:49

4Cl2+H2S+4H2O------------------------>8HCl+H2SO4

n(Cl2)=V(Cl)/Vm=17.92*10(-3)/22.4=0.0008 mol

n(H2SO4)=n(Cl2)/4=0.0008/4=0.0002 mol