Ta có :
\(\text{nCO2=0.02}\)
\(\text{Na2CO3+2HCl-->2NaCl+H2O+CO2}\)
0.02............0.04...........0.04..................0.02................(mol)
\(\text{a. nHCl=0.04}\)
\(\Rightarrow\text{CM(HCl)=0.04/0.02=2}\)
b. Sau phản ứng chỉ thu được 1 muối là NaCl
c. mNa2CO3=0.02*106=2.12
\(\Rightarrow\text{%mNa2CO3=2.12/5=42.4%}\)
\(\text{%mNaCl=57.6%}\)