$n_{H_2}=4,48/22,4=0,2mol$

$PTHH :$

$Zn+2HCl\to ZnCl_2+H_2↑$

$\text{a/Theo pt :}$

$n_{Zn}=n_{H_2}=0,2mol$

$⇒m_{Zn}=0,2.65=13g$

$\text{c/Theo pt :}$

$n_{HCl}=2.n_{H_2}=2.0,2=0,4mol$

$⇒m_{HCl}=0,4.36,5=14,6g$

$⇒C\%_{HCl}=\dfrac{14,6}{13+200-0,2.2}.100\%=6,87\%$

Đáp án:

 13g

7,3%

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
n{H_2} = \dfrac{V}{{22,4}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
nZn = n{H_2} = 0,2\,mol\\
mZn = n \times M = 0,2 \times 65 = 13g\\
c)\\
nHCl = 2n{H_2} = 0,4\,mol\\
mHCl = n \times M = 0,4 \times 36,5 = 14,6g\\
C\% HCl = \dfrac{{mct}}{{m{\rm{dd}}}} \times 100\%  = \dfrac{{14,6}}{{200}} \times 100\%  = 7,3\% 
\end{array}\)