Giải thích các bước giải:

\(\begin{array}{l}
A(Fe,S) \to B(Fe,S,FeS) \to C({H_2},{H_2}S)\\
B(Fe,S,FeS) \to (F{e_2}{O_3},S{O_2})
\end{array}\)

Bằng sơ đồ dường chéo, ta có:

\(\begin{array}{l}
\dfrac{{{n_{{H_2}}}}}{{{n_{{H_2}S}}}} = \dfrac{{34 - 26}}{{26 - 2}} = \dfrac{1}{3}\\
{n_{{H_2}}} + {n_{{H_2}S}} = V\\
 \to {n_{{H_2}}} = 0,25V \to {n_{{H_2}S}} = 0,75V
\end{array}\)

\(\begin{array}{l}
{n_{Fe}}(B) = {n_{{H_2}}} = 0,25V \to {m_{Fe}} = 14Vg\\
{n_{FeS}}(B) = {n_{{H_2}S}} = 0,75V \to {m_{FeS}} = 66Vg
\end{array}\)

Phản ứng của B với oxi:

\(\begin{array}{l}
4FeS + 7{O_2} \to 2F{e_2}{O_3} + 4S{O_2}\\
{n_{{O_2}}} = \dfrac{7}{4}{n_{FeS}} = 1,3125Vmol\\
S + {O_2} \to S{O_2}\\
{n_S} = {n_{{O_2}}} = amol\\
1,3125V + a = {V_1} \to a = {V_1} - 1,3125V\\
{m_S} = 32{V_1} - 42V\\
{m_B} = {m_{Fe}} + {m_S} + {m_{FeS}} = 14V + 32{V_1} - 42V + 66V = 38V + 32{V_1}
\end{array}\)

\(\begin{array}{l}
 \to \% {m_{Fe}} = \dfrac{{14V}}{{38V + 32{V_1}}} \times 100\% \\
\% {m_{FeS}} = \dfrac{{66V}}{{38V + 32{V_1}}} \times 100\% \\
\% {m_S} = \dfrac{{32{V_1} - 42V}}{{38V + 32{V_1}}} \times 100\% 
\end{array}\)

b,

\({n_S} = {V_1} - 1,3125V > 0 \to \dfrac{{{V_1}}}{V} > 1,3125\)