9.

Bảo toàn khối lượng:

$m_{Cl_2}=m_{muối}-m_{kl}=8,775-3,45=5,325(gam)$

$n_{Cl_2}=\dfrac{5,325}{71}=0,075(mol)$

$2A+Cl_2\xrightarrow{t^o}2ACl$

 0,15   0,075

$M_{A}=\dfrac{3,45}{0,15}=23(g/mol)$ 

A là Na.

10. Sửa đề 24g oxít kl

$m_{HCl}=21,9\times 10\text{%}=2,19(gam)$

$n_{HCl}=\dfrac{2,19}{36,5}=0,06(mol)$

$RO+2HCl→RCl_2+H_2O$

  0,03        0,06

$M_{RO}=\dfrac{24}{0,3}=80(g/mol)$

$M_R=M_{RO}-M_{R}=80-16=64(g/mol)$

Vậy R là Cu

Em tham khảo nha :

\(\begin{array}{l}
9)\\
2A + C{l_2} \to 2ACl\\
{n_A} = \dfrac{{3,45}}{{{M_A}}}mol\\
{n_{ACl}} = \dfrac{{8,775}}{{{M_A} + 35,5}}mol\\
{n_A} = {n_{ACl}}\\
\dfrac{{3,45}}{{{M_A}}} = \dfrac{{8,775}}{{{M_A} + 35,5}}\\
 \Rightarrow {M_A} = 23dvC\\
A:Natri(Na)\\
10)\\
RO + 2HCl \to RC{l_2} + {H_2}O\\
{n_{HCl}} = \dfrac{{21,9 \times 10\% }}{{36,5}} = 0,06mol\\
{n_{RO}} = \dfrac{{{n_{HCl}}}}{2} = 0,03mol\\
{M_{RO}} = \dfrac{{2,4}}{{0,03}} = 80dvC\\
{M_R} = 80 - 16 = 64dvC\\
R:\text{Đồng}(Cu)\\
CTHH:CuO
\end{array}\)