`Mg+2HCl->MgCl_2+H_2`

`Zn+2HCl->ZnCl_2+H_2`

a) `n_(Mg)=a;n_(Zn)=b`

`=>24a+65b=44,5`

`a+b=n_(H_2)=1`

`=>a=b=0,5(mol)`

`%m_(Mg)=(24.0,5)/(44,5)=26,97%`

`%m_(Zn)=100-26,97=73,03%`

b) `n_(HCl)=2n_(H_2)=2(mol)`

`=>m_(HCl)=(2.36,5)/(3,65).100=2000(g)`

`=>V=(m)/D=(2000)/(1,1)=1,818(l)`

`V_(ty)=1,818.20%+1,818=2,2725(l)`

c) `m_(ddsaupu)=m_(KL)+m_(axit)-m_(H_2)`

`=44,5+2000-2=2042,5(g)`

`C%_(Mg)=((24+71). 0,5)/(2042,5).100=2,36%`

`C%_(Zn)=(0,5.(71+65))/(2042,5).100=3,33%`

Đáp án:

\(\begin{array}{l}
a)\\
\% {m_{Mg}} = 26,97\% \\
\% {m_{Zn}} = 73,03\% \\
b){V_{{\rm{dd}}HCl(tt)}} = 2,275l\\
c)\\
C{\% _{MgC{l_2}}} = 2,33\% \\
C{\% _{ZnC{l_2}}} = 3,33\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{{H_2}}} = 1mol\\
a)
\end{array}\)

Gọi a và b lần lượt là số mol của Mg và Zn

\(\begin{array}{l}
\left\{ \begin{array}{l}
24a + 65b = 44,5\\
a + b = 1
\end{array} \right. \to \left\{ {a = b = 0,5} \right.\\
 \to {n_{Mg}} = {n_{Zn}} = 0,5mol\\
 \to \% {m_{Mg}} = \dfrac{{0,5 \times 24}}{{44,5}} \times 100\%  = 26,97\% \\
 \to \% {m_{Zn}} = 100\%  - 26,97\%  = 73,03\% 
\end{array}\)

\(\begin{array}{l}
b)\\
{n_{HCl}} = 2{n_{Mg}} + 2{n_{Zn}} = 2mol\\
 \to {m_{HCl}} = 73g\\
 \to {m_{{\rm{dd}}HCl}} = \dfrac{{73}}{{3,65\% }} \times 100\%  = 2000g\\
 \to {V_{{\rm{dd}}HCl}} = \dfrac{{2000}}{{1,1}} = 1818,2ml = 1,82l\\
 \to {V_{{\rm{dd}}HCl(tt)}} = \dfrac{{1,82 \times 125}}{{100}} = 2,275l
\end{array}\)

\(\begin{array}{l}
c)\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,5mol\\
 \to {m_{MgC{l_2}}} = 47,5g\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,5mol\\
 \to {m_{ZnC{l_2}}} = 68g\\
{m_{{\rm{dd}}}} = {m_{hỗnhợp}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 2042,5g\\
 \to C{\% _{MgC{l_2}}} = \dfrac{{47,5}}{{2042,5}} \times 100\%  = 2,33\% \\
 \to C{\% _{ZnC{l_2}}} = \dfrac{{68}}{{2042,5}} \times 100\%  = 3,33\% 
\end{array}\)