Bài tập về tính acid của H2SO4 loãng lớp 11 (có lời giải)

Đáp án đúng là: A

Trong X, chỉ có Fe tác dụng với dung dịch H$_{2}$SO$_{4}$ loãng.

$${n_{{H_2}}} = \frac{{2{,}479}}{{24{,}79}} = 0{,}1(mol)$$

$$\begin{array}{l}Fe\;\; + \;\;{H_2}S{O_4}\; \to {\rm{ }}FeS{O_4}\; + {\rm{ }}{H_2}\\0{,}1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \leftarrow \;\;\;\;\;\;\;0{,}1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol\;\\ \Rightarrow {m_{Fe}} = 0{,}1.56 = 5{,}6(g)\\ \Rightarrow {m_{Cu}} = 10 - 5{,}6 = 6{,}4(g).\end{array}$$

Đáp án đúng là: D

$$\begin{array}{l}{n_{{H_2}}} = \frac{{3{,}7185}}{{24{,}79}} = 0{,}15(mol)\\2A\;\; + \;\;n{H_2}S{O_4}\; \to {\rm{ }}\;{A_2}{\left( {S{O_4}} \right)_n}\; + {\rm{ }}n{H_2}\;{\rm{ }}\\\;\frac{{0{,}3}}{n}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \leftarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0{,}15\;\;\;\;\;\;\;\;\;\;\;(mol)\\ \Rightarrow {M_A} = \frac{{2{,}7}}{{\frac{{0{,}3}}{3}}} = 9n\end{array}$$

Đáp án đúng là: A

$${n_{{H_2}S{O_4}}} = 0{,}3.{\rm{ }}0{,}1 = 0{,}03{\rm{ }}\left( {mol} \right)$$

Vì phản ứng vừa đủ nên: ${n_{O{\rm{ }}(trong{\rm{ }}oxide)}}\; = {\rm{ }}{n_{{H_2}S{O_4}}}\; = 0{,}03{\rm{ }}\left( {mol} \right)$

$$\begin{array}{llllllllllllllllllll}{ \Rightarrow {\rm{ }}{m_{O(trong{\rm{ }}oxide)}}\; = {\rm{ }}0{,}03.{\rm{ }}16 = 0{,}48{\rm{ }}\left( g \right)}\\{ \Rightarrow {m_{KL}}\; = 2{,}81--0{,}48 = 2{,}33{\rm{ }}\left( g \right)}\end{array}$$

m$_{muối}$ $\; = {\rm{ }}{m_{KL}}\; + {\rm{ }}{m_{SO_4^{2 - }}}\; = 2{,}33 + 0{,}03.96 = 5{,}21{\rm{ }}\left( g \right).$

Đáp án đúng là: C

$${n_{{H_2}}} = \frac{{9{,}916}}{{24{,}79}}\; = 0{,}4{\rm{ }}mol$$

Bảo toàn nguyên tố H: ${n_{{H_2}}} = {n_{{H_2}SO{}_4}} = 0{,}4{\rm{ }}mol$

Bảo toàn khối lượng ta có:

$$\begin{array}{l}{m_{KL}}\; + {\rm{ }}{m_{{H_2}S{O_4}}}\; = {m_{{H_2}}}\; + {\rm{ }}{m_{muoi}}\;\\ \Leftrightarrow {\rm{ }}11{,}1 + 0{,}4.98 = 0{,}4.2 + m{\rm{ }} \Rightarrow {\rm{ }}m{\rm{ }} = {\rm{ 49{,}5(g)}}{\rm{.}}\end{array}$$

Đáp án đúng là: C

$${n_{Zn}}\; = \frac{{6{,}5}}{{65}} = 0{,}1{\rm{ }}mol$$

Zn + H$_{2}$SO$_{4}$ → ZnSO$_{4}$ + H$_{2}$

0,1 → 0,1

Þ m$_{Muối}$ = 0,1.(65 + 96) = 16,1 gam.

Đáp án đúng là: C

$${n_{{H_2}}} = \frac{{9{,}916}}{{24{,}79}}\; = {\rm{ }}0{,}4{\rm{ }}mol$$

Bảo toàn nguyên tố H: ${n_{{H_2}}} = {n_{{H_2}SO{}_4}} = 0{,}4{\rm{ }}mol$

Bảo toàn khối lượng ta có

$$\begin{array}{l}{m_{KL}}\; + {\rm{ }}{m_{{H_2}S{O_4}}}\; = {m_{{H_2}}}\; + {\rm{ }}{m_{muoi}}\;\\ \Leftrightarrow {\rm{ }}11{,}9 + 0{,}4.98 = 0{,}4.2 + m{\rm{ }} \Rightarrow {\rm{ }}m{\rm{ }} = {\rm{ }}50{,}3(g).\end{array}$$

Đáp án đúng là: D

Công thức oxide kim loại X là XO.

Giả sử có 1 mol XO

$\begin{matrix}\begin{matrix}\mathrm{XO}\ +\ {H}_{2}{\mathrm{SO}}_{4}\rightarrow \ {\mathrm{XSO}}_{4}+\ {H}_{2}O \\ 1\ \mathrm{mol}\rightarrow 1\ \mathrm{mol}\ 1\ \mathrm{mol}\end{matrix} \\ {m}_{{H}_{2}{\mathrm{SO}}_{4}}=98.1=98\ \mathrm{gam}\Rightarrow {m}_{\mathrm{dd}\ {H}_{2}{\mathrm{SO}}_{4}}=\frac{98.100}{10}=980(g)\end{matrix}$

Bảo toàn khối lượng ta có:

m$_{dd sau phản ứng}$= ${m_{dd{\rm{ }}{H_2}S{O_4}\;}} + {\rm{ }}{m_{XO\;}} = {\rm{ }}980{\rm{ }} + (X + 16).1 = 980 + X + 16{\rm{ (}}g)$

$$\Rightarrow C{\% _{dd{\rm{ }}{H_2}S{O_4}}} = \frac{{(X + 96).1}}{{980 + X + 16}}.100\% = 11{,}76\% \Rightarrow X = 24(Mg).$$

Đáp án đúng là: C

$${n_{{H_2}}} = \frac{{1{,}4874}}{{24{,}79}}\; = {\rm{ }}0{,}06{\rm{ }}mol$$

Bảo toàn nguyên tố H: ${n_{{H_2}}} = {n_{{H_2}SO{}_4}} = 0{,}06{\rm{ }}mol$

Bảo toàn khối lượng ta có:

$$\begin{array}{l}{m_{KL}}\; + {\rm{ }}{m_{{H_2}S{O_4}}}\; = {m_{{H_2}}}\; + {\rm{ }}{m_{muoi}}\;\\ \Leftrightarrow {\rm{ 3{,}22}} + 0{,}06.98 = 0{,}06.2 + m{\rm{ }} \Rightarrow m = {\rm{ 8{,}98(g)}}{\rm{.}}\end{array}$$

Đáp án đúng là: D

$$\begin{array}{l}{n_{{H_2}S{O_4}}} = 0{,}1.0{,}2 = 0{,}02(mol)\\{n_{NaOH}} = 0{,}2.0{,}1 = 0{,}02(mol)\\\begin{array}{llllllllllllllllllll}{2M\;\; + \;\;n{H_2}S{O_4}\; \to {\rm{ }}\;{M_2}{{\left( {S{O_4}} \right)}_n}\; + {\rm{ }}n{H_2}\;{\rm{ }}\left( 1 \right)}\\{\frac{{0{,}02}}{n}\;\;\; \leftarrow \;(0{,}02-0{,}01)}\\{{H_2}S{O_4}\;{\rm{ }} + {\rm{ }}2NaOH{\rm{ }} \to {\rm{ }}N{a_2}S{O_{4\;}}\; + 2{H_2}O\;\left( 2 \right)\;}\\{0{,}01\;\;\;\;\; \leftarrow \;\;\;\;\;\;0{,}02}\end{array}\\ \to M = \frac{m}{n} = \frac{{0{,}24}}{{\frac{{0{,}02}}{n}}} = 12n\end{array}$$

Đáp án đúng là: A

$${n_{{H_2}}} = \frac{{2{,}9748}}{{24{,}79}}\; = 0{,}12{\rm{ }}mol$$

Bảo toàn nguyên tố H: ${n_{{H_2}}} = {n_{{H_2}SO{}_4}} = 0{,}12{\rm{ }}mol$

Bảo toàn khối lượng ta có:

$${m_{KL}}\; + {\rm{ }}{m_{{H_2}S{O_4}}}\; = {m_{{H_2}}}\; + {\rm{ }}{m_{muoi}}\; \Leftrightarrow {\rm{ 6{,}05}} + 0{,}12.98 = 0{,}12.2 + m{\rm{ }} \Rightarrow m = {\rm{ 17{,}57(g)}}{\rm{.}}$$

Đáp án đúng là: A

Giả sử ${n_{{H_2}S{O_4}\;}} = {\rm{ }}1{\rm{ }}mol$

$$\begin{array}{llllllllllllllllllll}{M{{\left( {OH} \right)}_2}\; + {\rm{ }}{H_2}S{O_{4\;}} \to {\rm{ }}MS{O_4}\; + {\rm{ }}2{H_2}O}\\{1\;\;\;\;\;\;\;\;\;\;\;\; \leftarrow \;\;\;\;\;\;\;1\;\;\;\;\;\;\; \to \;\;\;\;\;\;\;\;1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {mol} \right)}\\{{m_{dd{\rm{ }}{H_2}S{O_4}}} = \frac{{1.98.100}}{{20}} = 490{\rm{ }}\left( {gam} \right)}\\\begin{array}{l} \to {\rm{ }}{m_{dd{\rm{ }}sau\;}} = 1.\left( {M + 34} \right) + 490 = M + 524(g)\\C{\% _{MS{O_4}}} = \frac{{M + 96}}{{M + 524}}.100\% = 27{,}21\% \\ \to M = 64(Cu)\end{array}\end{array}$$

Đáp án đúng là: B

$X\{\begin{matrix}F\text{e}:\ 0{,}2\ \mathrm{mol} \\ \mathrm{Mg}:\ 0{,}1\ \mathrm{mol}\end{matrix}\overset{+{H}_{2}{\mathrm{SO}}_{4}(L)}{\rightarrow }\ \{\begin{matrix}{\mathrm{FeSO}}_{4} \\ {\mathrm{MgSO}}_{4}\end{matrix}\overset{+\mathrm{NaOH}}{\rightarrow }↓\{\begin{matrix}\mathrm{Fe} \\ \mathrm{Mg}\end{matrix}\rightarrow _{\mathrm{kk}}^{{t}^{o}}\{\begin{matrix}{\mathrm{Fe}}_{2}{O}_{3}:\ 0{,}1\ \mathrm{mol}\ (\mathrm{BT}.\mathrm{Fe}) \\ \mathrm{MgO}:\ 0{,}1\ \mathrm{mol}\ (\mathrm{BT}.\ \mathrm{Mg})\end{matrix}$

⇒ m$_{rắn}$ = 0,1.160 + 0,1.40 = 20 gam.

a. Sai. Khí Y là H$_{2}$ không mùi.

b. Đúng. ${\text{n}}_{\text{M}\text{g}}\text{=}\ \frac{4{,}8}{24}\text{=}\ \text{0}\text{,}\text{2}\ \text{(}\text{m}\text{o}\text{l}\text{)}\text{​}\ \text{,}\ {\text{n}}_{{\text{H}}_{2}\text{S}{\text{O}}_{\text{4}}}\text{=}\ \text{0}\text{,}\text{2}\text{.}\text{1}\text{,}\text{5}\ \text{=}\ \text{0}\text{,}\text{3}\ \text{(}\text{m}\text{o}\text{l}\text{)}$

PTHH: Mg + H$_{2}$SO$_{4}$→ MgSO$_{4}$ + H$_{2}$

→ H$_{2}$SO$_{4}$ dư, tính theo Mg.

${\text{n}}_{{\text{H}}_{\text{2}}}\text{=}\ {\text{n}}_{\text{M}\text{g}}\text{=}\ \text{0}\text{,}\text{2}\ \text{(}\text{m}\text{o}\text{l}\text{)}\rightarrow {\text{V}}_{{\text{H}}_{\text{2}}}\text{=}\ 4{,}958\ (\text{L})$

c. Sai. H$_{2}$SO$_{4}$ dư

d. Đúng. $\text{n}{}_{\text{H}{}^{\text{+}}\text{d}\text{u}}\text{=}\ \text{2}\text{n}{}_{\text{H}{}_{\text{2}}\text{S}\text{O}{}_{\text{4}}\text{d}\text{u}}=0{,}2\ \text{m}\text{o}\text{l}\rightarrow [\text{H}{}^{+}]=\frac{0{,}2}{0{,}2}=1{,}0\ \text{M}\text{.}$

a. Đúng. Lớp oxide tan dần, sau đó lá kim loại cũng tan và có bọt khí xuất hiện.

H$_{2}$SO$_{4}$(aq) + ZnO(s) → ZnSO$_{4}$(aq) + H$_{2}$O(l)

H$_{2}$SO$_{4}$(aq) + Zn(s) → ZnSO$_{4}$(aq) + H$_{2}$(g)

b. Sai. Phản ứng giữa H$_{2}$SO$_{4}$ và CaCO$_{3}$ không có sự thay đổi số oxi hoá của nguyên tố nên không phải phản ứng oxi hoá khử: H$_{2}$SO$_{4}$(aq) + CaCO$_{3}$(s) → CaSO$_{4}$(s) + CO$_{2}$(g) + H$_{2}$O(1)

c. Đúng. H$_{2}$SO$_{4}$(aq) + 2NaHCO$_{3}$(s) → Na$_{2}$SO$_{4}$(aq) + 2CO$_{2}$(g) + 2H$_{2}$O(1)

d. Sai. Thí nghiệm (IV) không xuất hiện sủi bọt khí mà dung dịch thu được có vẩn đục:

H$_{2}$SO$_{4}$(aq) + Ca(OH)$_{2}$(aq) → CaSO$_{4}$(s) + 2H$_{2}$O(l).

a. Đúng. CuO là chất rắn màu đen, dung dịch CuSO$_{4}$ có màu xanh lam.

b. Sai. CuO + H$_{2}$SO$_{4}$→ CuSO$_{4}$ + H$_{2}$O

${\text{n}}_{\text{C}\text{u}\text{O}}\text{=}\ \frac{16}{80}\text{=}\ \text{0}\text{,}\text{2}\ \text{(}\text{m}\text{o}\text{l}\text{)}\text{​}\ \text{,}\ {\text{n}}_{{\text{H}}_{2}\text{S}{\text{O}}_{\text{4}}}\text{=}\ \frac{150.19{,}6\%}{98}\text{=}\ \text{0}\text{,}\text{3}\ \text{(}\text{m}\text{o}\text{l}\text{)}$→ CuO hết.

${\text{n}}_{{\text{H}}_{2}\text{S}{\text{O}}_{\text{4}}\ \text{p}\text{u}}\text{=}\ {\text{n}}_{\text{C}\text{u}\text{O}}=0{,}2\ \text{(}\text{m}\text{o}\text{l}\text{)}$

c. Đúng. ${\text{n}}_{\text{C}\text{u}\text{S}{\text{O}}_{\text{4}}}\text{=}\ {\text{n}}_{\text{C}\text{u}\text{O}}=0{,}2\ \text{(}\text{m}\text{o}\text{l}\text{)}$→ ${\text{m}}_{\text{C}\text{u}\text{S}{\text{O}}_{\text{4}}}\text{=}\ \text{0}\text{,}\text{2}\text{.}\text{1}\text{6}\text{0}\ \text{=}\ \text{3}\text{2}\ \ \text{(}\text{g}\text{)}$

d. Đúng. m$_{dd sau pư}$ = 16 + 150 = 166 (g). $\text{C}{\text{\%}}_{\text{C}\text{u}\text{S}{\text{O}}_{\text{4}}}\text{=}\ \frac{32}{166}\text{.}\text{1}\text{0}\text{0}\text{\%}\ \text{=}\ \text{1}\text{9}\text{,}\text{2}\text{8}\text{\%}$

a. Đúng. ${\text{n}}_{{\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}}\text{=}\ \text{0}\text{,}\text{1}\text{.}\text{0}\text{,}\text{2}\ \text{=}\ \text{0}\text{,}\text{0}\text{2}\ \text{(}\text{m}\text{o}\text{l}\text{)}\text{;}\ \ {\text{n}}_{\text{N}\text{a}\text{O}\text{H}}\ \text{=}\ \text{0}\text{,}\text{1}\text{.}\text{0}\text{,}\text{3}\ \text{=}\ \text{0}\text{,}\text{0}\text{3}\ \text{(}\text{m}\text{o}\text{l}\text{)}$

Pthh: H$_{2}$SO$_{4}$ + 2NaOH → Na$_{2}$SO$_{4}$ + H$_{2}$O

0,015 ← 0,03 (mol)

${\text{n}}_{{\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}\ \text{d}\text{u}}\text{=}\ \text{0}\text{,}\text{0}\text{2}-\text{0}\text{,}\text{0}\text{1}\text{5}\ \text{=}\ \text{0}\text{,}\text{0}\text{0}\text{5}\ \text{(}\text{m}\text{o}\text{l}\text{)}$

Vì H$_{2}$SO$_{4}$ dư nên dd Y có môi trường acid, làm quỳ tím hóa đỏ.

b. Sai. ${\text{n}}_{\text{O}{\text{H}}^{\text{-}}\ \text{p}\text{u}}\text{=}\ {\text{n}}_{\text{N}\text{a}\text{O}\text{H}}\ \text{=}\ \text{0}\text{,}\text{0}\text{3}\ \text{(}\text{m}\text{o}\text{l}\text{)}$

c. Đúng. ${\text{n}}_{{\text{H}}^{\text{+}}\text{d}\text{u}}\text{=}\ \text{2}{\text{n}}_{{\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}\ \text{d}\text{u}}\text{=}\ \text{0}\text{,}\text{0}\text{0}\text{5}\text{.}\text{2}\ \text{=}\ \text{0}\text{,}\text{0}\text{1}\ \text{(}\text{m}\text{o}\text{l}\text{)}$

d. Đúng. $\text{[}{\text{H}}^{+}]=\frac{0{,}01}{0{,}2}=0{,}05\ \text{M}\rightarrow \text{p}\text{H}\ \text{=}\ \text{-}\text{l}\text{o}\text{g}\text{(}\text{0}\text{,}\text{0}\text{5}\text{)}\approx \text{1}\text{,}\text{3}$

a. Đúng. Acid mạnh đẩy acid yếu hơn ra khỏi muối (Na$_{2}$CO$_{3}$+ H$_{2}$SO$_{4}$→ Na$_{2}$SO$_{4}$ + CO$_{2}$ + H$_{2}$O)

b. Đúng. ${\text{n}}_{\text{C}{\text{O}}_{\text{2}}}\text{=}\ {\text{n}}_{{\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}}\text{=}\ \text{0}\text{,}\text{2}\text{5}\text{.}\text{0}\text{,}\text{8}\ \text{=}\ \text{0}\text{,}\text{2}\ \text{(}\text{m}\text{o}\text{l}\text{)}$

c. Đúng. CO$_{2}$ + Ca(OH)$_{2}$ → CaCO$_{3}$ + H$_{2}$O

Theo pthh: ${\text{n}}_{\text{C}\text{a}\text{C}{\text{O}}_{\text{3}}}\text{=}\ {\text{n}}_{\text{C}{\text{O}}_{\text{2}}}\text{=}\ \text{0}\text{,}\text{2}\ \text{(}\text{m}\text{o}\text{l}\text{)}\rightarrow {\text{m}}_{\text{C}\text{a}\text{C}{\text{O}}_{\text{3}}}=0{,}2.100=20\ (\text{g})$

d. Sai. Thay dung dịch Na$_{2}$CO$_{3}$ bằng bột BaCO$_{3}$ thì phản ứng vẫn có khí thoát ra:

BaCO$_{3}$ + H$_{2}$SO$_{4}$→ BaSO$_{4}$ + CO$_{2}$ + H$_{2}$O

Đáp án: 7,6.

${\text{n}}_{{\text{H}}_{\text{2}}}=\frac{1{,}2395}{24{,}79}=0{,}05\ \text{(}\text{m}\text{o}\text{l}\text{)}$

PTHH: Fe + H$_{2}$SO$_{4}$ → FeSO$_{4}$ + H$_{2}$

0,05 ← 0,05

a = ${\text{m}}_{\text{F}\text{e}\text{S}{\text{O}}_{\text{4}}}\text{=}\ \text{0}\text{,}\text{0}\text{5}\text{.}\text{1}\text{5}\text{2}\ \text{=}\ \text{7}\text{,}\text{6}\ \ \text{(}\text{g}\text{)}$

Đáp án: 62.

Đặt ${\text{n}}_{\text{Z}\text{n}\text{S}{\text{O}}_{\text{4}}}\text{=}\ \text{x}\ \text{,}\ {\text{n}}_{\text{C}\text{u}\text{S}{\text{O}}_{\text{4}}}\text{=}\ \text{y}\ \text{(}\text{m}\text{o}\text{l}\text{)}$

→ ${\text{n}}_{\text{Z}\text{n}}\text{=}\ \text{x}\ \text{,}\ {\text{n}}_{\text{C}\text{u}\text{O}}\text{=}\ \text{y}\ \text{(}\text{m}\text{o}\text{l}\text{)}$

Ta có: 65x + 80y = 21

${\text{n}}_{\text{S}{\text{O}}_{\text{4}}^{\text{2}\text{-}}}\text{=}\ {\text{n}}_{{\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}}\rightarrow \text{x}\ \text{+}\ \text{y}\ \text{=}\ \text{0}\text{,}\text{6}\text{.}\text{0}\text{,}\text{5}$

→ x = 0,2 mol, y = 0,1 mol

%m$_{Zn}$ = $\frac{0{,}2.65}{21}.100\%=62\%$

Đáp án: 4.

Có 4 kim loại tác dụng được H$_{2}$SO$_{4}$ loãng là: Mg, Al, Zn, Fe.

Đáp án: 7,437.

${\text{n}}_{\text{A}\text{l}}\text{=}\ \frac{5{,}4}{27}\text{=}\ \text{0}\text{,}\text{2}\ \text{(}\text{m}\text{o}\text{l}\text{)}$

PTHH: 2Al + 3H$_{2}$SO$_{4}$→ Al$_{2}$(SO$_{4}$)$_{3}$ + 3H$_{2}$

0,2 → 0,3 (mol)

${\text{V}}_{{\text{H}}_{\text{2}}}\text{=}\ \text{0}\text{,}\text{3}\text{.}\text{2}\text{4}\text{,}\text{7}\text{9}\ \text{=}\ \text{7}\text{,}\text{4}\text{3}\text{7}\ \text{(}\text{L}\text{)}$

Đáp án: 4.

Bao gồm:

(2) Fe(OH)$_{2}$ + H$_{2}$SO$_{4 (loãng)}$→ FeSO$_{4}$ + 2H$_{2}$O

(3) Ca + H$_{2}$SO$_{4 (loãng)}$ → CaSO$_{4}$ + H$_{2}$

(4) H$_{2}$SO$_{4 (loãng)}$ + CuO → CuSO$_{4}$ + H$_{2}$O

(5) BaCl$_{2}$ + H$_{2}$SO$_{4 (loãng)}$ → BaSO$_{4}$↓ + 2HCl