Question

6. 500g topitura nacl p=90% supusa electrolizei. se cere:
ec. react la electroliza
V la 227 grde si 1 atm
mHCl rez la un randament al electrolizei de 80%.

Answer 1

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 !!!!!