Đáp án:

32,A

33,C

34,D

35,B

36,B

37,C

38,C

39,B

40,D

Giải thích các bước giải:

\(\begin{array}{l}
32,\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
{n_{CaC{O_3}}} = 0,4mol\\
 \to {n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,4mol\\
 \to V = {V_{C{O_2}}} = 8,96l
\end{array}\)

\(\begin{array}{l}
33,\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = 0,01mol\\
 \to {n_{HCl}} = 6{n_{F{e_2}{O_3}}} = 0,06mol\\
 \to V = {V_{HCl}} = 0,12l = 120ml
\end{array}\)

\(\begin{array}{l}
34,\\
BaC{l_2} + N{a_2}S{O_4} \to BaS{O_4} + 2NaCl\\
{n_{BaC{l_2}}} = 0,02mol\\
{n_{N{a_2}S{O_4}}} = 0,015mol\\
 \to {n_{BaC{l_2}}} > {n_{N{a_2}S{O_4}}}
\end{array}\)

Suy ra \(BaC{l_2}\) dư

\( \to {n_{BaS{O_4}}} = {n_{N{a_2}S{O_4}}} = 0,015mol\)

\(\begin{array}{l}
35,\\
3NaOH + FeC{l_3} \to Fe{(OH)_3} + 3NaCl\\
{n_{NaOH}} = 0,12mol\\
{n_{FeC{l_3}}} = 0,02mol\\
 \to \dfrac{{{n_{NaOH}}}}{3} > {n_{FeC{l_3}}}
\end{array}\)

Suy ra NaOH dư

\(\begin{array}{l}
 \to {n_{Fe{{(OH)}_3}}} = {n_{FeC{l_3}}} = 0,02mol\\
 \to a = {m_{Fe{{(OH)}_3}}} = 2,14g
\end{array}\)

\(\begin{array}{l}
36,\\
\dfrac{{{n_{KOH}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,4}}{{0,1}} = 4\\
2KOH + C{O_2} \to {K_2}C{O_3} + {H_2}O\\
 \to \dfrac{{{n_{KOH}}}}{2} > {n_{C{O_2}}}
\end{array}\)

Suy ra KOH dư 

\( \to {n_{{K_2}C{O_3}}} = {n_{C{O_2}}} = 0,1mol\)

\(\begin{array}{l}
37,\\
\dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,3}}{{0,1}} = 3\\
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O\\
 \to \dfrac{{{n_{NaOH}}}}{2} > {n_{C{O_2}}}
\end{array}\)

Suy ra NaOH dư

\(\begin{array}{l}
 \to {n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,1mol\\
 \to {m_{N{a_2}C{O_3}}} = 10,6g
\end{array}\)

38,

\(\begin{array}{l}
Al + 3AgN{O_3} \to Al{(N{O_3})_3} + 3Ag\\
{n_{AgN{O_3}}} = 0,03mol\\
 \to {n_{Al}}phản ứng= \dfrac{1}{3}{n_{AgN{O_3}}} = 0,01mol\\
 \to {n_{Ag}} = {n_{AgN{O_3}}} = 0,03mol\\
 \to {m_{tăng}} = 0,03 \times 108 - 0,01 \times 27 = 2,97g
\end{array}\)

\(\begin{array}{l}
39,\\
Cu + 2AgN{O_3} \to Cu{(N{O_3})_2} + 2Ag
\end{array}\)

Gọi a là số mol của Cu phản ứng 

\( \to {n_{Ag}} = 2{n_{Cu}} = 2amol\)

\(\begin{array}{l}
 \to {m_{tăng}} = {m_{Ag}} - {m_{Cu}} = 2a \times 108 - 64a = 1,52g\\
 \to a = 0,01mol\\
 \to {n_{AgN{O_3}}} = 2{n_{Cu}} = 0,02mol\\
 \to x = C{M_{AgN{O_3}}} = \dfrac{{0,02}}{{0,2}} = 0,1M
\end{array}\)

40,

\(\begin{array}{l}
{n_{{H_2}}} = 0,4mol\\
 \to {n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,4mol\\
 \to {n_{S{O_4}^{2 - }}} = {n_{{H_2}S{O_4}}} = 0,4mol\\
 \to m = {m_{Kim loại}} + {m_{S{O_4}^{2 - }}} = 12 + 0,4 \times 96 = 50,4g
\end{array}\)