Đáp án:

 1) $Zn$

2) $\% {m_{Zn}} = 61,61\% ;\% {m_{ZnO}} = 38,39\% $

3) 17,84%

Giải thích các bước giải:

 1) ${n_{{H_2}S{O_4}}} = 0,2.1.5 = 0,3mol;{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol$

$2M + n{H_2}S{O_4} \to {M_2}{(S{O_4})_n} + n{H_2}$  (1)

${M_2}{O_n} + n{H_2}S{O_4} \to {M_2}{(S{O_4})_n} + n{H_2}O$ (2)

Từ (1) $ \Rightarrow {n_M} = \dfrac{2}{n}.{n_{{H_2}}} = \dfrac{{0,4}}{n}$

$ \Rightarrow {n_{{H_2}S{O_4}(1)}} = {n_{{H_2}}} = 0,2mol \Rightarrow {n_{{H_2}S{O_4}(2)}} = 0,3 - 0,2 = 0,1mol$

Từ (2) $ \Rightarrow {n_{{M_2}{O_n}}} = \dfrac{1}{n}{n_{{H_2}S{O_4}(2)}} = \dfrac{{0,1}}{n}$

${m_A} = {m_M} + {m_{{M_2}{O_n}}}$

$ \Leftrightarrow \dfrac{{0,4}}{n}.M + \dfrac{{0,1}}{n}(2M + 16n) = 21,1$

$ \Rightarrow M = 32,5n$

⇒ n = 2; M = 65 

⇒ M là $Zn$

2) ${n_{Zn}} = \dfrac{{0,4}}{2} = 0,2mol;{n_{ZnO}} = {n_{{H_2}S{O_4}}} = 0,1mol$

$\begin{gathered}
   \Rightarrow \% {m_{Zn}} = \dfrac{{0,2.65}}{{21,1}}.100\%  = 61,61\%  \hfill \\
   \Rightarrow \% {m_{ZnO}} = 100 - 61,61 = 38,39\%  \hfill \\ 
\end{gathered} $

3) ${m_{dd{H_2}S{O_4}}} = V.D = 1,25.200 = 250g$

Bảo toàn khối lượng: ${m_A} + {m_{dd{H_2}S{O_4}}} = {m_{ddB}} + {m_{{H_2}}}$

$ \Rightarrow {m_{ddB}} = 21,1 + 250 - 0,2.2 = 270,7g$

$\sum {{n_{ZnS{O_4}}} = {n_{Zn}} + {n_{ZnO}}}  = 0,3mol$

$ \Rightarrow C{\% _{ZnS{O_4}}} = \dfrac{{0,3.161}}{{270,7}}.100\%  = 17,84\% $

a,Ta có: hỗn hợp A gồm 21,1 g $M$ và $M_2O_x$

$2M+xH_2SO_4→M_2(SO_4)_x+xH_2$(1)

 $M_2O_x+xH_2SO_4→M_2(SO_4)_x+xH_2O$(2)

$n_{H_2SO_4}=0,3(mol)$
$n_{H_2}=0,2(mol)$⇒$n_M=\dfrac{0,4}{x}(mol)$

Ta có:$n_{H_2SO_4(2)}=0,1(mol)$

⇒$n_{M_2O_x}=\dfrac{0,1}{x}(mol)$ 

⇒$\dfrac{M.0,4}{x} +\dfrac{0,1(2M+16x)}{x} =21,1$

⇒$0,4M+0,2M+1,6x=21,1x$⇒$0,6M=19,5x⇒M=32,5x$
⇒Với x=2 thì M=65⇒ M là Zn

b,Ta có: $n_{Zn}=0,2(mol)$⇒$m_{Zn}=13(g)$

$n_{ZnO}=0,1(mol)$⇒$m_{ZnO}=8,1(g)$

⇒%$Zn=61,6$%⇒%$ZnO=38,4$%

c,$m_{H_2SO_4}=1,25.200=250(g)$

Ta có: $m_{ddsp ư}=250+21,1-0,4=270,7(g)$

$n_{ZnSO_4}=0,2+0,1=0,3(mol)$⇒$m_{ZnSO_4}=48,3(g)$

⇒C%$ZnSO_4=17,84$%