Corpul uman contine 8% azot. Considerand masa organismului uman de 70 kg calculeaza: a)cantitatea de azot existenta in corpul uman b)cantitatea de solutie de NH3 60% ce s-ar putea obtine din azotul corpului uman

a)Masa de azot ce se gaseste in corpul uman este:

[tex]m=\frac{8}{100}*70kg=5,6kg=5600g[/tex]

Cantitatea de azot existenta in  corpul uman este:

[tex]n=\frac{m}{A_N}=400moli ; A_N=14g/mol[/tex]


   2*17g    28g  
b)2NH3=N2+3H2
      x     5600g

[tex]M_{NH_3}=A_N+3A_H=14+3*1=17g/mol[/tex]

[tex]M_{N_2}=2A_N=28g/mol[/tex]

[tex]x=\frac{5600*2*17}{28}=6800g \ NH_3 \ pur[/tex]

[tex]c=\frac{m_d*100}{m_s}=\textgreater m_s=\frac{m_d*100}{c}=11333,33g[/tex]

[tex]M_{mediu}=\frac{60*M_{NH_3}}{100}+\frac{40*M_{H_2O}}{100}=17,4g/mol(M_{H_2O}=2A_H+2A_O=18g/mol)[/tex]

[tex]n_1=\frac{m_s}{M_{mediu}}=651,34moli[/tex]