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

a. Sai. Khí Y là H₂ 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 + H₂SO₄→ MgSO₄ + H₂

→ H₂SO₄ 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{)}\to {\text{V}}_{{\text{H}}_{\text{2}}}\text{=}4,958(\text{L})

c. Sai. H₂SO₄ 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}\to [\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.

H₂SO₄(aq) + ZnO(s) → ZnSO₄(aq) + H₂O(l)

H₂SO₄(aq) + Zn(s) → ZnSO₄(aq) + H₂(g)

b. Sai. Phản ứng giữa H₂SO₄ và CaCO₃ không có sự thay đổi số oxi hoá của nguyên tố nên không phải phản ứng oxi hoá khử: H₂SO₄(aq) + CaCO₃(s) → CaSO₄(s) + CO₂(g) + H₂O(1)

c. Đúng. H₂SO₄(aq) + 2NaHCO₃(s) → Na₂SO₄(aq) + 2CO₂(g) + 2H₂O(1)

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

H₂SO₄(aq) + Ca(OH)₂(aq) → CaSO₄(s) + 2H₂O(l).

a. Đúng. CuO là chất rắn màu đen, dung dịch CuSO₄ có màu xanh lam.

b. Sai. CuO + H₂SO₄ → CuSO₄ + H₂O

{\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\mathrm{\%}}{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: H₂SO₄ + 2NaOH → Na₂SO₄ + H₂O

            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 H₂SO₄ 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}\to \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 (Na₂CO₃ + H₂SO₄ → Na₂SO₄ + CO₂ + H₂O)

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. CO₂ + Ca(OH)₂ → CaCO₃ + H₂O

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{)}\to {\text{m}}_{\text{C}\text{a}\text{C}{\text{O}}_{\text{3}}}=0,2.100=20(\text{g})

d. Sai. Thay dung dịch Na₂CO₃ bằng bột BaCO₃ thì phản ứng vẫn có khí thoát ra:

BaCO₃ + H₂SO₄ → BaSO₄ + CO₂ + H₂O

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 + H₂SO₄ → FeSO₄ + H₂

                                      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}}}\to \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\mathrm{\%}=62\mathrm{\%}

Đáp án: 4.

Có 4 kim loại tác dụng được H₂SO₄ 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 + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

            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 + H₂SO_{4 (loãng)} → FeSO₄ + 2H₂O

(3) Ca + H₂SO_{4 (loãng)} → CaSO₄ + H₂

(4) H₂SO_{4 (loãng)} + CuO → CuSO₄ + H₂O

(5) BaCl₂ + H₂SO_{4 (loãng)} → BaSO₄↓ + 2HCl