NaCl(topitura)-------> Na⁺ + Cl⁻
K(-) Na⁺+e--.> Na ori2
A(+) 2Cl⁻-2e---> Cl2
m,impur=500gNaCl
p/100= m,pur/m,impur----> m,pur= 450g NaCl pur
M(miu)= (23gNa+35,5gCl)/mol= 58,5g/mol
n,mol= m/M; n= 450mol/58,5g/mol=>7,69mol NaCl
2mol.........................1mol
2NaCl-----> 2Na + Cl2
7,69mol......................n= 3,845 mol Cl2
pV= nRT; 1xV= 3,845x0,082x(227+273)...................calculeaza V,in LITRI
.....1mol...........2mol
H2 +Cl2--_-> 2HCl
3,845.... 7,69molHCl
m= nxM---> m=7,69mol(1gH+35,5gCl)/mol.................calculeaza !!!!!
