Question

Reactioneaza 400g argint de puritate 80% cu acidul azotic, solutie de concentratie 63%. Se cere:
a) Masa de argint pur.
b) Masa dizolvantului (acid azotic) din sol.
c) Masa sol. acid azotic.
d) Nr. moli sare obtinut.
e) Volumul de gaz rezultat.

Answer 1

m Ag = 400g
p = 80%
C = 63% sol. HNO₃

Ag + 2HNO₃⇒ AgNO₃ + NO₂↑ + H₂O

a) p = m pur/m impur×100 ⇒ m pur =  80×400/100 = 320g Ag

b) 108g Ag....................2×63g HNO₃
320g Ag.........................xg HNO₃

x = (2×63×320)/108 = 373,33g HNO₃ (md)

c) C = md/ms×100 ⇒ ms = md×100/C = 37333/63 = 529,59g sol. HNO₃

d) 108g  Ag..................1 mol AgNO₃
320g Ag.......................x moli AgNO₃

x = 320×1/108 = 2,962 moli AgNO₃

e) n AgNO₃ = n NO₂ = 2,962 moli

V NO₂ = 22,4*2,962 = 66,37 L