nHCl=0,2.0,2=0,04
Ta có: \(nH^+=nOH^-\)
\(\Rightarrow nOH^-=0,04\)
\(\Rightarrow V\)NaOH=\(\dfrac{0,04}{0,2}\)=0,2(l)
PT: 2HCl+ CaCO3\(\rightarrow\)CaCl2+H2O+CO2
nCaCO3=\(\dfrac{0,04}{2}\)=0,02\(\Rightarrow\)mCaCO3=0,02.100=2
nCO2=0,02
\(\Rightarrow\)VCO2=0,02.22,4=0,448(l)