Sa koliko atoma kiseonika se jedini 448 mL azota (normalni uslovi) pri gradjenu anhidrida azotaste kiseline?
2N2+3O2--->2N2O3 n(N2)=448/1000/22.4 mol=0.02mol n(N2):n(O2)=2:3 n(O2)=0.03mol n(O)=2*n(O2)=0.06mol N(O)=0.06*6*10^23=3.6*10^22
powered by Drupal
2N2+3O2--->2N2O3 n(N2)=448/10
2N2+3O2--->2N2O3
n(N2)=448/1000/22.4 mol=0.02mol
n(N2):n(O2)=2:3
n(O2)=0.03mol
n(O)=2*n(O2)=0.06mol
N(O)=0.06*6*10^23=3.6*10^22