p= m,px100/m,i---> m,p= 1500x95/100kg= 1425kg NaCl pur
2NaCl---> 2Na⁺ +2Cl⁻
KATOD; 2Na⁺+2e---> 2Na
ANOD : 2Cl⁻-2e---Cl2
2NaCl-> 2Na⁺ + 2Cl⁻====> 2Na + Cl2
din 2mol NaCl se obtin 2 molNa si 1 molCl2
sau 2x58,5g,,,,,,,,,,,,,,,,,,,,2x23g 1mol M,NaCl=58,5g/mol
1425kg............................x........................y
x=560kg Na -masa teoretica, cu randament 100%
y=12,18 kmolCl2, numar de moli teoretic,cu randament 100%
80/100= m,pr/560-----------> calculeaza m,prNa
80/100= niu,pr/12,18---> niu,pr=9,74kmolCl2
Cl2 + H2--->2HCl
1mol............2mol
9,74kmol........niu=19,48kmol HCl
m=niuxM= 19.48x36,5 kg=711kg-HCl
nu stiu cat de bine este ,pentru ca nu sunt clare datele din textul tau !
