Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Fe}} = 36,84\% \\
\% {m_{FeO}} = 63,16\% \\
b)\\
{C_\% }FeC{l_2} = 11,93\% \\
c)\\
m = 138,25g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,15\,mol\\
{m_{Fe}} = 0,15 \times 56 = 8,4g\\
{m_{FeO}} = 22,8 - 8,4 = 14,4g\\
\% {m_{Fe}} = \dfrac{{8,4}}{{22,8}} \times 100\% = 36,84\% \\
\% {m_{FeO}} = 100 - 36,84 = 63,16\% \\
b)\\
{n_{FeO}} = \dfrac{{14,4}}{{72}} = 0,2\,mol\\
{n_{HCl}} = 0,15 \times 2 + 0,2 \times 2 = 0,7\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,7 \times 36,5}}{{7,3\% }} = 350g\\
{m_{{\rm{dd}}spu}} = 22,8 + 350 - 0,15 \times 2 = 372,5g\\
{n_{FeC{l_2}}} = 0,15 + 0,2 = 0,35\,mol\\
{C_\% }FeC{l_2} = \dfrac{{0,35 \times 127}}{{372,5}} \times 100\% = 11,93\% \\
c)\\
2AgN{O_3} + FeC{l_2} \to Fe{(N{O_3})_2} + 2AgCl\\
Fe{(N{O_3})_2} + AgN{O_3} \to Fe{(N{O_3})_3} + Ag\\
{n_{AgCl}} = 2{n_{FeC{l_2}}} = 0,7\,mol\\
{n_{Ag}} = {n_{Fe{{(N{O_3})}_2}}} = 0,35\,mol\\
m = 0,7 \times 143,5 + 0,35 \times 108 = 138,25g
\end{array}\)
