$\text{Câu 3:}$

$n_{Cl_2}=\dfrac{58.8-16.2}{35.5 \times 2}=0.6(mol)$

$n_{O_2}=\dfrac{63.6-58.8}{32}=0.15(mol)$

Bảo toàn $e:$

$n \times n_{R}=2n_{Cl_2}+4n_{O_2}$

`=>`$M_R=\dfrac{16.2n}{1.8}$

Tại $n=1$`=>`$M_R=9$

Tại $n=2$`=>`$M_R=18$

Tại $n=3$`=>`$M_R=27$`=>`$Al$

Bảo toàn $e:$

$3n_{Al/X}=4n_{O_2}$

`=>`$n_{Al/X}=0.2(mol)$

`=>`$m_{Al}=5.4(gam)$

$1)S$

$2)S$

$3) Đ$

$4)S$

$\text{Câu 4:}$

$n_{Mn O_2}=0.36(mol)$

Bảo toàn $e:$

$2n_{Cl_2}=2n_{MnO_2}$

`=>`$n_{X}=n_{Cl_2}=0.36(mol)$

`=>`$V_X=8.9244(L)$

$n_{KOH}=2n_{Cl_2}=0.72(mol)$

`=>`$V_{KOH}=\dfrac{0.72}{1.5}=0.48(L)=480(mL)$

$2KOH+Cl_2 \to KCl+KClO+H_2O$

$n_{KCl}=n_{Cl_2}=036(mol)$

$C_M(KCl)=\dfrac{0.36}{1.5}=0.75(M)$

$1)Đ$

$2)Đ$

$3)S$

$4)Đ$