Giải thích các bước giải:

1,

\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
{n_{{H_2}}} = 0,1mol = {n_{Mg}} \to {m_{Mg}} = 2,4g\\
 \to {m_{MgO}} = 3,6g \to {n_{MgO}} = 0,09mol\\
 \to \% {m_{Mg}} = \dfrac{{2,4}}{6} \times 100\%  = 40\% \\
 \to \% {m_{MgO}} = 60\% \\
{n_{HCl}} = 2{n_{Mg}} + 2{n_{MgO}} = 0,38mol \to {m_{HCl}} = 13,87g\\
 \to {m_{HCl{\rm{dd}}}} = \dfrac{{13,87 \times 100}}{{20}} = 69,35g\\
 \to {V_{HCl{\rm{dd}}}} = \dfrac{m}{D} = 63,05ml
\end{array}\)

2,

Gọi CT của oxit có dạng \({M_2}O\)

\(\begin{array}{l}
{M_2}O + {H_2}S{O_4} \to {M_2}S{O_4} + {H_2}O\\
{n_{{M_2}O}} = {n_{{M_2}S{O_4}}}\\
 \to \dfrac{{6,2}}{{2M + 16}} = \dfrac{{14,2}}{{2M + 96}}\\
 \to M = 23\\
 \to Na
\end{array}\)

Vậy A là Natri

Đáp án:

`1)\ %m_{Mg}=40%\ ; %m_{MgO}=60%\ ; V_{dd\ HCl}=63,045(ml)`

`2)\ A: Natri\ (Na)`

Giải thích các bước giải:

`1)`

`a)`

`Pt: Mg+2HCl->MgCl_2+H_2`

       `MgO+2HCl->MgCl_2+H_2O`

`n_{H_2}={2,24}/{22,4}=0,1(mol)`

`n_{Mg}=n_{H_2}=0,1(mol)`

`%m_{Mg}={0,1.24}/{6}.100%=40%`

`%m_{MgO}=100%-40%`

`b)`

`n_{MgO}={6-(0,1.24)}/{40}=0,09(mol)`

`\Sigman_{HCl}=2(n_{Mg}+n_{MgO})=0,38(mol)`

`m_{dd\ HCl}={0,38.36,5}/{20%}=69,35(g)`

`V_{dd\ HCl}={69,35}/{1,1}=63,045(ml)`

 `2)`

`O x it: R_2O`

`Pt: R_2O+H_2SO_4->R_2SO_4+H_2O`

`Ta\ co: {6,2}/{M_R.2+16}={14,2}/{M_R.2+96}`

`=>M_R=23`$(g/mol)$

`A: Natri\ (Na)`