Question

sa se determine valenta azotului in oxizii care contin:
a)30,45% N si 69,57% O
b) 36,84% N si 63,16% O.
Multumesc!

Answer 1

a.  30,45% N , 69,57% O  - dioxid de azot

$N:\ \frac{30.45}{14}=2.175\ |:2.175\implies N\rightarrow1\\\\O:\ \frac{69.57}{16}=4.348\ |:2.175\implies O\rightarrow2\\\\\implies F.M.:\ NO_2$

N.O.:   N -> +4 , O -> -2   [⇒1×(+4) + 2×(-2) = 4 - 4 = 0]

b. 36,84% N , 63,16% O  - trioxid de diazot

$N:\ \frac{36.84}{14}=2.631\ |:2.631\implies N\rightarrow1\ |\cdot2\implies N\rightarrow2\\\\O:\ \frac{63.16}{16}=3.947\ |:2.631\implies O\rightarrow1.5\ |\cdot2\implies O\rightarrow3\\\\\implies F.M.:\ N_2O_3$

N.O.:   N -> +3 , O -> -2   [⇒2×(+3) + 3×(-2) = 6 - 6 = 0]

Obs(1): oxigenul are N.O. întotdeauna -2, cu excepția apei oxigenate (H₂O₂), unde are -1.

Obs(2): suma N.O. dintr-un compus trebuie să fie 0.