a. Ecuația reacției chimice:
0.4 x
CaCO₃ + 2HCl -----: CaCl₂ + CO₂ + H₂O
1 1
b.
[tex]V_s=200\ mL\\C_M=2\ M\\C_M=\frac{n}{V_s}\implies n=C_M\ {\cdot}\ V_s=2{\cdot}200=400\ mmoli=0.4\ moli\ CaCO_3\\\\M_{CaCO_3}=40+12+(3{\cdot}16)=100\ g/mol\\n=\frac{m}{M}\implies m=n{\cdot}M=0.4{\cdot}100=40\ g\ CaCO_3[/tex]
c.
[tex]n_{CO_2}=x=0.4\ moli\\V_m=22.4\ L\\n=\frac{V}{V_m}\implies V=n\ {\cdot}\ V_m=0.4\ {\cdot}\ 22.4=8.96\ L\ CO_2[/tex]
