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ce cantitate de NaOH neutralizeaza 1 mol de HCL?

Răspuns :

NaOH + HCl = NaCl + HOH
[tex]M_{NaOH} [/tex] = 23+16+1= 40 g/mol 
[tex] M_{HCl} [/tex] = 1+35,5=36,5 g/mol
n=[tex] \frac{m}{M} [/tex] => m=1*36,5=36,5 g HCl
[tex] \frac{x}{40} [/tex]=[tex] \frac{36,5}{36,5} [/tex] => x=[tex] \frac{36,5*40}{36,5} [/tex] 
=> x=40 g NaOH