Question

va rog ,ajutor exercitiul 3!!​

Answer 1

$V_{N_2}=4.48\ m^3\\V_M=22.4\ dm^3\\n=\frac{V}{V_M}=\frac{4.48}{22.4}=0.2\ kmoli\ N_2$

a.

1 mol N₂ ................. NA molecule (NA = 6.023×10²³)

0.2kmoli N₂ .............. x molecule      ⇒  x = 1.2046×10²⁶ molecule

b.

$M_{N_2}=2*14=28\ g/mol\\n=\frac{m}{M}\implies\ m=n*M=0.2*28=5.6\ kg\ N_2$

c.

      x          0.2kmoli

    3H₂     +     N₂     ----->     2NH₃

      3               1

$M_{H_2}=2*1=2\ g/mol\\n_{H_2}=x=0.6\ kmoli\\m_{H_2}=n*M=0.6*2=1.2\ kg\\V_{H_2}=n*22.4=0.6*22.4=13.44\ m^3$