PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\)
a) Ta có: \(n_{Na_2CO_3}=\frac{53}{106}=0,5\left(mol\right)\)
\(\Rightarrow n_{NaCl}=1mol\) \(\Rightarrow m_{NaCl}=1\cdot58,5=58,5\left(g\right)\)
b) Theo PTHH: \(n_{CO_2}=n_{Na_2CO_3}=0,5mol\)
\(\Rightarrow V_{CO_2}=0,5\cdot22,4=11,2\left(l\right)\)