PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

Ta có: \(n_{Al}=\dfrac{3,24}{27}=0,12\left(mol\right)\)

a, \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,09\left(mol\right)\) \(\Rightarrow V_{O_2}=0,09.22,4=2,016\left(l\right)\)

b, \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,06\left(mol\right)\) \(\Rightarrow m_{Al_2O_3}=0,06.102=6,12\left(g\right)\)

c, \(V_{kk}=\dfrac{2,016}{21\%}=9,6\left(l\right)\)

d, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,18\left(mol\right)\)

\(\Rightarrow m_{KMnO_4}=0,18.158=28,44\left(g\right)\)

Bạn xem lời giải ở đây nhé.

https://hoc24.vn/cau-hoi/cho-324-g-al-tac-dung-voi-oxi-vua-du-th-duoc-al2o3-a-tinh-vo2-b-tinh-m-al2o3-c-trong-vkk-can-dung-biet-vo2-21-vkk-d-tinh-khoi-luong-kmno.7651142171785

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

Theo PT:  \(n_{Ca\left(OH\right)_2}=n_{CaCO_3}=n_{CO_2}=0,25\left(mol\right)\)

a, \(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)

b, \(m_{CaCO_3}=0,25.100=25\left(g\right)\)

c, \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

Theo PT: \(n_{HCl}=2n_{Ca\left(OH\right)_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{20\%}=91,25\left(g\right)\)

Đáp án C

6FeSO₄+K₂Cr₂O₇+7H₂SO₄→3Fe₂(SO₄)₃+Cr₂(SO₄)₃+K₂SO₄+7H₂O

→nK₂Cr₂O₇=0,1mol→mK₂Cr₂O₇=29,4

Chọn D.

Đáp án D

K₂Cr₂O₇ + 0,06 mol FeSO₄ + H₂SO₄ → Cr₂SO₄ + Fe₂(SO₄)₃ + K₂SO₄ + H₂O
• Theo bảo toàn e: 6 × n_K2Cr2O7 = 2 × n_FeSO4 → n_K2Cr2O7 = 1 × 0,6 : 6 = 0,1 mol
→ m_K2Cr2O7 = 0,1 × 294 = 29,4 gam

Chọn đáp án D

6 F e S O 4 + K 2 C r 2 O 7 + 7 H 2 S O 4 → 3 F e S O 4 3 + C r 2 S O 4 3 + K 2 S O 4 + 7 H 2 O

→ n K 2 C r 2 O 7 = 0 , 1 m o l → m K 2 C r 2 O 7 = 29 , 4