Đáp án:
Bạn tham khảo lời giải nhé !
Giải thích các bước giải:
\(\begin{array}{l}
2/\\
3CO + F{e_2}{O_3} \to 2Fe + 3C{O_2}\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O
\end{array}\)
\(\begin{array}{l}
a.\\
{n_{CaC{O_3}}} = 0,03mol\\
\to {n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,03mol\\
\to {n_{F{e_2}{O_3}}} = \dfrac{1}{3}{n_{C{O_2}}} = 0,01mol\\
\to {m_{F{e_2}{O_3}}} = 1,6g
\end{array}\)
\(\begin{array}{l}
b.\\
{n_{CO}} = {n_{C{O_2}}} = 0,03mol\\
\to {V_{CO}} = 0,672l
\end{array}\)
\(\begin{array}{l}
3/\\
F{e_2}{O_3} + 3CO \to 2Fe + 3C{O_2}\\
{n_{F{e_2}{O_3}}} = 0,1mol\\
\to {n_{CO}} = {n_{C{O_2}}} = 3{n_{F{e_2}{O_3}}} = 0,3mol\\
\to {V_{CO}} = 6,72l
\end{array}\)
\(\begin{array}{l}
{m_{{\rm{dd}}}}_{KOH} = \dfrac{{99,12}}{{1,17}} = 84,72g\\
\to {m_{KOH}} = \dfrac{{84,72 \times 20\% }}{{100\% }} = 17g\\
\to {n_{KOH}} = 0,3mol\\
\to \dfrac{{{n_{KOH}}}}{{{n_{C{O_2}}}}} = 1
\end{array}\)
=> Tạo 1 muối: \(KHC{O_3}\)
\(\begin{array}{l}
C{O_2} + KOH \to KHC{O_3}\\
{n_{KHC{O_3}}} = {n_{KOH}} = 0,3mol\\
\to {m_{KHC{O_3}}} = 30g
\end{array}\)
\(\begin{array}{l}
4/\\
F{e_2}{O_3} + 3CO \to 2Fe + 3C{O_2}\\
CuO + CO \to Cu + C{O_2}\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O
\end{array}\)
\(\begin{array}{l}
{n_{CaC{O_3}}} = 0,04mol\\
\to {n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,04mol\\
\to {n_{CO}} = {n_{C{O_2}}} = 0,04mol\\
\to {V_{CO}} = 0,896l
\end{array}\)
