Đáp án:

\(\begin{array}{l}
31,D\\
32,A\\
33,A\\
34,D\\
35,A\\
36,A\\
37,B\\
38,A\\
39,B\\
40,C
\end{array}\)

Giải thích các bước giải:

31,

\(\begin{array}{l}
{m_{{H_2}S{O_4}}} = 98g\\
 \to {m_{dd}} = 98 + 18 = 116g\\
 \to C{\% _{{H_2}S{O_4}}} = \dfrac{{98}}{{116}} \times 100\%  = 84,48\% 
\end{array}\)

32,

\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}O\\
{n_{Fe}} = 0,2mol\\
{n_{HCl}} = 0,3mol\\
 \to {n_{Fe}} > \dfrac{{{n_{HCl}}}}{2}
\end{array}\)

Suy ra Fe dư 

\(\begin{array}{l}
 \to {n_{{H_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,15mol\\
 \to {m_{{H_2}}} = 0,3g
\end{array}\)

33,

\(\begin{array}{l}
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{{H_2}}} = 0,3mol\\
 \to {n_{Fe}} = {n_{{H_2}}} = 0,3mol\\
 \to {m_{Fe}} = 16,8g
\end{array}\)

34,

\(\begin{array}{l}
S{O_3} + {H_2}O \to {H_2}S{O_4}\\
{n_{S{O_3}}} = 0,1mol\\
 \to {n_{{H_2}S{O_4}}} = {n_{S{O_3}}} = 0,1mol\\
 \to C{M_{{H_2}S{O_4}}} = \dfrac{{0,1}}{{0,4}} = 0,25M
\end{array}\)

35,

\(\begin{array}{l}
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
{n_{Na}} = 0,2mol\\
 \to {n_{NaOH}} = {n_{Na}} = 0,2mol\\
 \to C{M_{NaOH}} = \dfrac{{0,2}}{{0,5}} = 0,4M
\end{array}\)

36,

\(\begin{array}{l}
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
{n_{Na}} = 0,2mol\\
 \to {n_{{H_2}}} = \dfrac{1}{2}{n_{Na}} = 0,1mol\\
 \to {V_{{H_2}}} = 2,24l
\end{array}\)

37,

\(\begin{array}{l}
{P_2}{O_5} + 3{H_2}O \to 2{H_3}P{O_4}\\
{n_{{H_3}P{O_4}}} = 2{n_{{P_2}{O_5}}} = 0,2mol\\
 \to {m_{{H_3}P{O_4}}} = 19,6g\\
 \to C{\% _{{H_3}P{O_4}}} = \dfrac{{19,6}}{{0,1 \times 142 + 500}} \times 100\%  = 3,81\% 
\end{array}\)

38,

\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = 0,1mol\\
 \to {n_{HCl}} = 2{n_{Fe}} = 0,2mol\\
 \to C{\% _{HCl}} = \dfrac{{0,2 \times 36,5}}{{400}} \times 100\%  = 1,825\% 
\end{array}\)

39,

\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = 0,2mol\\
 \to {n_{{H_2}}} = {n_{Fe}} = 0,2mol\\
 \to {V_{{H_2}}} = 4,48l
\end{array}\)

40,

\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = 0,2mol\\
 \to {n_{FeC{l_2}}} = {n_{Fe}} = 0,2mol\\
 \to {n_{{H_2}}} = {n_{Fe}} = 0,2mol\\
 \to {m_{dd}} = 11,2 + 400 - 0,2 \times 2 = 410,8g\\
 \to C{\% _{FeC{l_2}}} = \dfrac{{0,2 \times 127}}{{410,8}} \times 100\%  = 6,18\% 
\end{array}\)

Đáp án:

Câu 31:

$m_{H_{2}SO_{4}}=1 × 98= 98 (g)$

⇒ $m_{dd_{H_{2}SO_{4}}}=98 + 18= 116 (g)$

⇒ $\text{C%$H_{2}SO_{4}=\dfrac{98}{116}$ × 100%= 84,48%}$

⇒ Chọn $D$

Câu 32:

$Fe + 2HCl→FeCl_{2} + H_{2}↑$

Ta có: $n_{HCl}=3 × 0,1= 0,3 (mol)$

Lập tỉ lệ: $\dfrac{0,2}{1} > \dfrac{0,3}{2}$

⇒ $Fe$ dư, tính theo $n_{HCl}$

Theo PT: $n_{H_{2}}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2} × 0,3= 0,15 (mol)$

⇒ $V_{H_{2}}=0,15 × 2= 0,3 (g)$

⇒ Chọn $A$

Câu 33: 

$Fe + H_{2}SO_{4}→FeSO_{4} + H_{2}↑$

Ta có: $n_{H_{2}}=\dfrac{6,72}{22,4}= 0,3 (mol)$

Theo PT: $n_{Fe}=n_{H_{2}}= 0,3 (mol)$

⇒ $m_{Fe}=0,3 × 56= 16,8 (g)$

⇒ Chọn $A$

Câu 34:

$SO_{3} + H_{2}O→H_{2}SO_{4}$

Ta có: $n_{SO_{3}}=\dfrac{8}{80}= 0,1 (mol)$

Theo PT: $n_{H_{2}SO_{4}}=n_{SO_{3}}= 0,1 (mol)$

⇒ $C_{M_{H_{2}SO_{4}}}=\dfrac{0,1}{0,4}= 0,25 (M)$

⇒ Chọn $D$

Câu 35:

$2Na + 2H_{2}O→2NaOH + H_{2}↑$

Ta có: $n_{Na}=\dfrac{4,6}{23}= 0,2 (mol)$

Theo PT: $n_{NaOH}=n_{Na}= 0,2 (mol)$

⇒ $C_{M_{NaOH}}=\dfrac{0,2}{0,5}= 0,4 (M)$

⇒ Chọn $B$

Câu 36:

$2Na + 2H_{2}O→2NaOH + H_{2}↑$

Ta có: $n_{Na}=\dfrac{4,6}{23}= 0,2 (mol)$

Theo PT: $n_{H_{2}}=\dfrac{1}{2}n_{Na}=\dfrac{1}{2} × 0,2= 0,1 (mol)$

⇒ $V_{H_{2}}=0,1 × 22,4= 2,24 (l)$

⇒ Chọn $A$

Câu 37: 

$P_{2}O_{5} + 3H_{2}O→2H_{3}PO_{4}$

Ta có: $m_{P_{2}O_{5}}=0,1 × 142= 14,2 (g)$

Theo PT: $n_{H_{3}PO_{4}}=2n_{P_{2}O_{5}}=2 × 0,1= 0,2 (mol)$

⇒ $m_{H_{3}PO_{4}}=0,2 × 98= 19,6 (g)$

$m_{\text{dd sau pứ}}=14,2 + 500= 514,2 (g)$

⇒ $\text{C%$H_{3}PO_{4}=\dfrac{19,6}{514,2}$ × 100%= 3,81%}$

⇒ Chọn $B$

Câu 38:

$Fe + 2HCl→FeCl_{2} + H_{2}↑$

Ta có: $n_{Fe}=\dfrac{5,6}{56}= 0,1 (mol)$

Theo PT: $n_{HCl}=2n_{Fe}=2 × 0,1= 0,2 (mol)$

⇒ $m_{HCl}=0,2 × 36,5= 7,3 (g)$

⇒ $\text{C%HCl=$\dfrac{7,3}{400}$ × 100%= 1,825%}$

⇒ Chọn $A$

Câu 39: 

$Fe + 2HCl→FeCl_{2} + H_{2}↑$

Ta có: $n_{Fe}=\dfrac{11,2}{56}= 0,2 (mol)$

Theo PT: $n_{Fe}=n_{H_{2}}= 0,2 (mol)$

⇒ $V_{H_{2}}=0,2 × 22,4= 4,48 (l)$

⇒ Chọn $B$

Câu 40: 

$Fe + 2HCl→FeCl_{2} + H_{2}↑$

Ta có: $n_{Fe}=\dfrac{11,2}{56}= 0,2 (mol)$

Theo PT: $n_{H_{2}}=n_{Fe}= 0,2 (mol)$

⇒ $m_{H_{2}}=0,2 × 2= 0,4 (g)$

⇒ $m_{\text{dd sau pứ}}=(11,2 + 400) - 0,4= 410,8 (g)$

Theo PT: $n_{FeCl_{2}}=n_{Fe}= 0,2 (mol)$

⇒ $m_{FeCl_{2}}=0,2 × 127= 25,4 (g)$

⇒ $\text{C%$FeCl_{2}=\dfrac{25,4}{410,8}$ × 100%= 6,18%}$

⇒ Chọn $C$