MgCO3+2HCl->MgCl2+H2O+CO2
0,25--------0,5
n MgCO3=\(\dfrac{21}{84}\)=0,25 mol
=>VHCl=\(\dfrac{0,5}{2}\)=0,25 l=250ml
=>B
\(n_{MgCO_3}=\dfrac{21}{84}=0,25mol\)
\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)
0,25 0,5 ( mol )
\(V_{HCl}=\dfrac{0,5}{2}=0,25l\)
=> Chọn B