Metali

magdalena's picture

Smesa NaOH i KOH mase 3.58g u reakciji sa HCL daje 5.04g hlorida ovih metala.Koliko ima KOH u smesi?

milicaradenkovic's picture

NaOH+HCl-------------------->NaCl+H2O

KOH+HCl------------------->KCl+H2O

m(hlorida)=m(KCl)+m(NaCl)

n(NaCl)=n(NaOH)

n(KCl)=n(KOH)

5.04=n(KCl)*Mr+n(NaCl)*Mr

m(smese)=m(NaOH)+m(KOH)

5.04=n(KOH)*74.5+n(NaOH)*58.5

3.58=n(NaOH)*40+n(KOH)*56

n(NaOH)=(5.04-n(KOH)*74.5)/58.5

3.58=((5.04-n(KOH)*74.5)/58.5)*40+n(KOH)*56

3.58=3.446+5.06*n(KOH)

0.134=5.06*n(KOH)

n(KOH)=0.134/5.06=0.0265 mol

m(KOH)=n(KOH)*Mr=0.0265*56=1.484 g

m(NaOH)=3.58-m(KOH)=3.58-1.484=2.096 g

 

magdalena's picture

hvala