Ta có: \(m_{HCl}=200.7,3\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
PT: \(K_2CO_3+2HCl\rightarrow2KCl+CO_2+H_2O\)
Theo PT: \(n_{CO_2}=n_{K_2CO_3}=\dfrac{1}{2}n_{HCl}=0,2\left(mol\right)\)
a, VCO2 = 0,2.24,79 = 4,958 (l)
b, \(C_{M_{K_2CO_3}}=\dfrac{0,2}{0,15}=\dfrac{4}{3}\left(M\right)\)