Anhidrid hipohloraste kiseline

Submitted by Danica88 on Sat, 03/15/2014 - 12:13
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Sa koliko atoma kiseonika se jedini 224 ml hlora. merenog pri normalnim uslovima, pri građenju anhidrida hipohloraste kiseline?

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milicaradenkovic - Sat, 03/15/2014 - 13:05

2Cl2+O2--------------------->2Cl2O

n(Cl2)=V(Cl2)/Vm=0.224/22.4=0.01 mol

n(O2):n(Cl2)=1:2

n(O2)=1*n(Cl2)/2=1*0.01/2=0.005 mol

n(O)=2*n(O2)=2*0.005=0.01 mol

N(O)=n(O)*Na=0.01*6*10(23)=6*10(21)

 

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Danica88 - Sat, 03/15/2014 - 19:15

Hvala puno! :)