a. \({\text{a}}{\text{. FeO}} + CO \to Fe + C{O_2}\)

\({\text{F}}{{\text{e}}_2}{O_3} + 3CO \to 2Fe + 3C{O_2}\)

\({\text{C}}{{\text{O}}_2} + 2NaOH \to N{a_2}C{O_3} + {H_2}O\)

\(N{a_2}C{O_3} + Ca{(OH)_2} \to CaC{O_3} + 2NaOH\)

\({n_{CaC{O_3}}} = \frac{{4,6}}{{100}} = 0,046{\text{ }}mol\)

\(Theo{\text{ PTHH: }}{{\text{n}}_{{\text{C}}{{\text{O}}_2}}} = {n_{N{a_2}C{O_3}}} = {n_{CaC{O_3}}} = 0,046{\text{ mol}}\)

\({{\text{n}}_{O/oxit}} = {{\text{n}}_{{\text{C}}{{\text{O}}_2}}} = 0,046{\text{ mol}}\)

\({{\text{m}}_A} = {m_B} + {m_O} = 4,784{\text{ }} + {\text{ }}0,046{\text{ x 16  =  5,52 gam}}\)

Gọi x, y lần lượt là số mol \({\text{FeO; F}}{{\text{e}}_2}{O_3}\)

Ta có hệ phương trình:

72x + 160y = 5,52

x + y = 0,04

Giải hệ ta có: x = 0,01; y = 0,03

\(% {m_{FeO}} = \frac{{0,01{\text{ x 72}}}}{{5,52}}{\text{ x 100  =  13,04% }}\)

\(% {m_{F{e_2}{O_3}}} = {\text{ 100  -  13,04  =  86,96% }}\)

b. Hỗn hợp rắn B gồm: \({\text{Fe; FeO; F}}{{\text{e}}_2}{O_3};{\text{ F}}{{\text{e}}_3}{O_4}\)

\({\text{Fe}} + 2HCl \to FeC{l_2} + {H_2}\)

\({\text{FeO}} + 2HCl \to FeC{l_2} + {H_2}O\)

\({\text{F}}{{\text{e}}_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\)

\({\text{F}}{{\text{e}}_3}{O_4} + 8HCl \to FeC{l_2} + 2FeC{l_3} + 4{H_2}O\)

\(Cu + 2FeC{l_3} \to 2FeC{l_2} + CuC{l_2}\)