Đáp án:

\(\begin{array}{l}
a)\\
\% {m_{Zn}} = 44,52\% \\
\% {m_{ZnO}} = 55,48\% \\
b)\\
{D_{{\rm{dd}}HCl}} = 1,19(g/ml)
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
ZnO + 2HCl \to ZnC{l_2} + {H_2}O\\
ZnC{l_2} + 2AgN{O_3} \to Zn{(N{O_3})_2} + 2AgCl\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{Zn}} = {n_{{H_2}}} = 0,1\,mol\\
{n_{AgCl}} = \dfrac{{57,4}}{{143,5}} = 0,4\,mol \Rightarrow {n_{ZnC{l_2}}} = \dfrac{{0,4}}{2} = 0,2\,mol\\
{n_{ZnO}} = 0,2 - 0,1 = 0,1\,mol\\
\% {m_{Zn}} = \dfrac{{0,1 \times 65}}{{0,1 \times 65 + 0,1 \times 81}} \times 100\%  = 44,52\% \\
\% {m_{ZnO}} = 100 - 44,52 = 55,48\% \\
b)\\
{n_{HCl}} = 0,1 \times 2 + 0,1 \times 2 = 0,4\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,4 \times 36,5}}{{3,65\% }} = 400g\\
{D_{{\rm{dd}}HCl}} = \dfrac{{400}}{{336}} = 1,19(g/ml)
\end{array}\)