Đáp án:

Giải thích các bước giải:

Em tham khảo nha :

\(\begin{array}{l}
1)\\
a)\\
A + 2HCl \to AC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_A} = {n_{{H_2}}} = 0,2mol\\
{M_A} = \dfrac{{4,8}}{{0,2}} = 24dvC\\
 \Rightarrow A:Magie(Mg)\\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,4mol\\
C{\% _{HCl}} = \dfrac{{0,4 \times 36,5}}{{200}} \times 100\%  = 7,3\% \\
2)\\
a)\\
Ba{(OH)_2} + {H_2}S{O_4} \to BaS{O_4} + 2{H_2}O\\
Zn{(OH)_2} + {H_2}S{O_4} \to ZnS{O_4} + 2{H_2}O\\
b)\\
{m_{{H_2}S{O_4}}} = \dfrac{{200 \times 9,8}}{{100}} = 19,6g\\
{n_{{H_2}S{O_4}}} = \dfrac{{19,6}}{{98}} = 0,2mol\\
hh:Ba{(OH)_2}(amol),Zn{(OH)_2}(b\,mol)\\
\left\{ \begin{array}{l}
171a + 99b = 27\\
a + b = 0,2
\end{array} \right.\\
 \Rightarrow a = 0,1;b = 0,1\\
{m_{Ba{{(OH)}_2}}} = 0,1 \times 171 = 17,1g\\
{m_{Zn{{(OH)}_2}}} = 0,1 \times 99 = 9,9g\\
c)\\
{n_{BaS{O_4}}} = {n_{Ba{{(OH)}_2}}} = 0,1mol\\
C{\% _{BaS{O_4}}} = \dfrac{{0,1 \times 233}}{{27 + 200}} \times 100\%  = 10,26\% \\
{n_{ZnS{O_4}}} = {n_{Zn{{(OH)}_2}}} = 0,1mol\\
{m_{ZnS{O_4}}} = \dfrac{{0,1 \times 161}}{{27 + 200}} \times 100\%  = 7,1\% 
\end{array}\)