Đáp án:

\(\begin{array}{l}
a)\\
m = 10,9g\\
b)\\
{m_{AgCl}} = 21,525g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
ZnO + 2HCl \to ZnC{l_2} + {H_2}O\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{HCl}} = 0,3 \times 1 = 0,3mol\\
{n_{{H_2}}} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,05mol\\
{m_{Fe}} = 0,05 \times 56 = 2,8g\\
{n_{ZnO}} = \dfrac{{{n_{HCl}} - 2{n_{Fe}}}}{2} = \dfrac{{0,3 - 0,1}}{2} = 0,1mol\\
{m_{ZnO}} = 0,1 \times 81 = 8,1g\\
m = 2,8 + 8,1 = 10,9g\\
b)\\
ZnC{l_2} + 2AgN{O_3} \to Zn{(N{O_3})_2} + 2AgCl\\
FeC{l_2} + 2AgN{O_3} \to Fe{(N{O_3})_2} + 2AgCl\\
{n_{ZnC{l_2}}} = {n_{ZnO}} = 0,1mol\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,05mol\\
{n_{AgCl}} = {n_{ZnC{l_2}}} + {n_{FeC{l_2}}} = 0,15mol\\
{m_{AgCl}} = 0,15 \times 143,5 = 21,525g
\end{array}\)