$PTHH:KClO_3+6HCl\to KCl+3Cl_2\uparrow+3H_2O(1)$

$K+HCl\to KCl+0,5H_2(2)$

$n_{K}=\dfrac{4,8}{39}=\dfrac{8}{65}(mol)$

Theo PT(2): $n_{HCl}=n_K=\dfrac{8}{65}(mol)$

Theo PT(1): $n_{KClO_3}=6n_{HCl}=\dfrac{48}{65}(mol)$

Và $n_{Cl_2}=0,5n_{HCl}=\dfrac{4}{65}(mol)$

$\Rightarrow V_{Cl_2}=\dfrac{4}{65}.22,4\approx 1,38(lít)$

$m_{KClO_3}=\dfrac{48}{65}.122,5\approx 90,46(g)$

Vậy $m=90,46;V=1,38$

$n_{K}=\dfrac{4,8}{39}=\dfrac{8}{65}(mol)\\ PTHH:\\ KClO_3+6HCl\to KCl+3Cl_2+3H_2O\\ 2K+2HCl\to 2KCl+H_2\\ \xrightarrow{\text{Từ PTHH}}\ n_{HCl}=n_K=\dfrac{8}{65}(mol)\\ \xrightarrow{\text{Từ PTHH}}\ \begin{cases}n_{KClO_3}=\dfrac{1}{6}.n_{HCl}=\dfrac{4}{195}(mol)\\ n_{Cl_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{4}{65}(mol)\end{cases}\\ m=\dfrac{4}{195}.122,5=2,51(g)\\ V=\dfrac{4}{65}.22,4=1,38(l)$