Ecuația reacției de neutralizare:
2,4moli x=2,4moli
NaOH + HCl ------------> NaCl + HOH
1mol 1mol
Rezolvarea problemei:
[tex]C_{M_{NaOH}}=2*1,5=3M \\ V_{s_{NaOH}}=800\ mL[/tex]
[tex]C_M=\frac{n}{V_s}[/tex] ⇒[tex]n=C_M*V_s=3*800=2400\ mmoli=2,4\ moli\ NaOH[/tex]
[tex]C_{M_{HCl}}=1,5M \\ n_{HCl}=x=2,4\ moli[/tex]
[tex]C_M=\frac{n}{V_s}[/tex] ⇒ [tex]V_s=\frac{2,4}{1,5}=1,6\ L=1600\ mL\ HCl[/tex]
