$CaO$ + $2HCl$ -> $CaCl_{2}$ + $H_{2}O$
Có : $n_{CaO}$ = $\frac{m_{CaO}}{M_{CaO}}$ = $\frac{5,6}{40+ 16}$ = $\frac{5,6}{56}$ = 0,1 mol
Theo PTHH : $n_{HCl}$ = $2n_{CaO}$
=> $n_{HCl}$ = 2 . 0,1 = 0,2 mol
Có : $V_{HCl}$ = $\frac{n_{HCl}}{CM_{HCl}}$ = $\frac{0,2}{1 }$ = 0,2 l