Đáp án:
21,D
22,B
23,C
24,C
Giải thích các bước giải:
21,
Gọi a và b là số mol của Fe và Cu
\(\begin{array}{l}
Fe + CuS{O_4} \to FeS{O_4} + Cu\\
\left\{ \begin{array}{l}
56a + 64b = 10\\
64a + 64b = 11
\end{array} \right.\\
\to a = 0,125 \to b = 0,05\\
\to \% {m_{Fe}} = \dfrac{{0,125 \times 56}}{{10}} \times 100\% = 70\% \\
\to \% {m_{Cu}} = 30\%
\end{array}\)
22,
\(\begin{array}{l}
{R_2}{O_3} + 3{H_2}S{O_4} \to {R_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{{R_2}{{(S{O_4})}_3}}} = {n_{{R_2}{O_3}}} = \dfrac{{10,2}}{{2R + 48}}mol\\
\to C{\% _{{R_2}{{(S{O_4})}_3}}} = \dfrac{{(2R + 288) \times \dfrac{{10,2}}{{2R + 48}}}}{{10,2 + 331,8}} \times 100 = 10\\
\to R = 27 \to A{l_2}{O_3}
\end{array}\)
23,
\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
CO + O \to C{O_2}\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,1mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,075mol\\
\to {n_O} = {n_{C{O_2}}} = 0,1mol\\
\to \dfrac{{{n_{Fe}}}}{{{n_O}}} = \dfrac{3}{4} \to F{e_3}{O_4}
\end{array}\)
24,
\(\begin{array}{l}
{C_2}{H_5}OH + Na \to {C_2}{H_5}ONa + \dfrac{1}{2}{H_2}\\
{V_{{C_2}{H_5}OH}} = 92 \times 100 = 9200ml\\
\to {m_{{C_2}{H_5}OH}} = 9200 \times 0,8 = 7360g\\
\to {n_{{C_2}{H_5}OH}} = 160mol\\
\to {n_{{H_2}}} = \dfrac{1}{2}{n_{{C_2}{H_5}OH}} = 80mol\\
\to {V_{{H_2}}} = 1792ml = 1,792l
\end{array}\)
