Question

500g Na2CO3 de concentratie 25% reactioneaza cu HCl, obtinundu-se NaCl,
dioxid de carbon si apa. Se cer urmatoarele:
A) cantitatea de HCl utilizat in reactie
B) cantitatea de dioxid de carbon​

Answer 1

Ecuația reacției chimice:

   1.18               x                                     y

Na₂CO₃   +   2HCl   ----->   2NaCl   +   CO₂   +   H₂O

     1                 2                                      1

Rezolvare:

$m_s_{Na_2CO_3}=500\ g\\C=25\%\\\\C=\frac{m_d}{m_s}{\cdot}100\implies m_d=\frac{m_s{\cdot}C}{100}=\frac{500{\cdot}25}{100}=125\ g\ Na_2CO_3\\\\M=2A_{Na}+A_C+3A_O=(2{\cdot}23)+(12)+(3{\cdot}16)=46+12+48=106\ g/mol\\\\n=\frac{m}{M}=\frac{125}{106}=1.18\ moli\ Na_2CO_3$

$n_{HCl}=x=2.36\ moli\\M_{HCl}=A_H+A_{Cl}=1+35.5=36.5\ g/mol\\\\n=\frac{m}{M}\implies m=n\ {\cdot}\ M=2.36\ {\cdot}\ 36.5=86.14\ g$

$n_{CO_2}=y=1.18\ moli\\M_{CO_2}=A_C+2A_O=12+(2{\cdot}16)=12+32=44\ g/mol\\\\n=\frac{m}{M}\implies m=n\ {\cdot}\ M=1.18\ {\cdot}\ 44=51.92\ g$