PTHH: \(CuSO_4+2NaOH\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

Ta có: \(\left\{{}\begin{matrix}n_{CuSO_4}=0,08\cdot3,5=0,28\left(mol\right)\\n_{NaOH}=0,12\cdot1,5=0,18\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,28}{1}>\dfrac{0,18}{2}\) \(\Rightarrow\) CuSO₄ còn dư

\(\Rightarrow\left\{{}\begin{matrix}n_{Na_2SO_4}=0,09\left(mol\right)=n_{Cu\left(OH\right)_2}\\n_{CuSO_4\left(dư\right)}=0,19\left(mol\right)\\\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Na_2SO_4}=0,09\cdot142=12,78\left(g\right)\\m_{Cu\left(OH\right)_2}=0,09\cdot98=8,82\left(g\right)\\m_{CuSO_4\left(dư\right)}=0,19\cdot160=30,4\left(g\right)\end{matrix}\right.\)

Mặt khác: \(\left\{{}\begin{matrix}C_{M_{Na_2SO_4}}=\dfrac{0,09}{0,08+0,12}=0,45\left(M\right)\\C_{M_{CuSO_4\left(dư\right)}}=\dfrac{0,19}{0,08+0,12}=0,95\left(M\right)\end{matrix}\right.\)

\(a,HCl+NaOH\rightarrow NaCl+H_2O\\ b,n_{HCl}=\dfrac{3,65}{36,5}=0,1mol\\ n_{NaOH}=0,2.1=0,2mol\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,2}{1}\Rightarrow NaOH.dư\\ n_{NaCl}=n_{NaOH,pư}=n_{HCl}=0,1mol\\ m_{NaOH,dư}=\left(0,2-0,1\right).40=4g\\ c,m_{NaCl}=0,1.58,5=5,85g\)

Ta có: \(n_{HCl}=0,2.1=0,2\left(mol\right)\)

\(n_{NaOH}=0,3.2=0,6\left(mol\right)\)

PT: \(HCl+NaOH\rightarrow NaCl+H_2O\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,6}{1}\), ta được NaOH dư.

Theo PT: \(n_{NaOH\left(pư\right)}=n_{HCl}=0,2\left(mol\right)\)

\(\Rightarrow n_{NaOH\left(dư\right)}=0,6-0,2=0,4\left(mol\right)\)

\(\Rightarrow m_{NaOH\left(dư\right)}=0,4.40=16\left(g\right)\)

- Quỳ tím hóa xanh do NaOH dư.

a) \(n_{NaOH}=\dfrac{60.11,2\%}{40}=0,168\left(mol\right)\)

PTHH: \(2NaOH+MgCl_2\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\)

            0,168---->0,084----->0,084

b) \(m_{kt}=m_{Mg\left(OH\right)_2}=0,084.58=4,872\left(g\right)\)

c) \(C\%_{MgCl_2}=\dfrac{0,084.95}{190}.100\%=4,2\%\)

Ban oi   cho minh hoi la 40 lay o dau a ?

\(n_{BaCl_2}=\dfrac{208.15\%}{208}=0,15\left(mol\right)\\ n_{H_2SO_4}=\dfrac{150.19,6\%}{98}=0,3\left(mol\right)\\ BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ Vì:\dfrac{0,15}{1}< \dfrac{0,3}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(p.ứ\right)}=n_{BaSO_4}=n_{BaCl_2}=0,15\left(mol\right)\\ n_{HCl}=2.0,15=0,3\left(mol\right)\\ n_{H_2SO_4\left(dư\right)}=0,13-0,15=0,15\left(mol\right)\\ m_{HCl}=0,3.36,5=10,95\left(g\right)\\ m_{BaSO_4}=233.0,15=34,95\left(g\right)\\ m_{H_2SO_4\left(dư\right)}=0,15.98=14,7\left(g\right)\\ m_{ddsau}=208+150-34,95=323,05\left(g\right)\\ C\%_{ddHCl}=\dfrac{10,95}{323,05}.100\approx3,39\%\)

\(C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{14,7}{323,05}.100\approx4,55\%\)

a,\(m_{BaCl_2}=208.15\%=31,2\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{31,2}{208}=0,15\left(mol\right)\)

\(m_{H_2SO_4}=150.19,6\%=29,4\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\)

PTHH: BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

Mol:      0,15       0,15          0,15        0,3

Ta có: \(\dfrac{0,15}{1}< \dfrac{0,3}{1}\) ⇒ BaCl₂ hết, H₂SO₄ dư

\(m_{H_2SO_4dư}=\left(0,3-0,15\right).98=14,7\left(g\right)\)

b, \(m_{BaSO_4}=0,15.233=34,95\left(g\right)\)

  \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)

Đáp án C

Bảo toàn e : 2n_Cu = 2n_SO2 => n_SO2 = 0,2 mol

n_NaOH = 0,4 mol = 2n_SO2

=> Chỉ có phản ứng : 2NaOH + SO₂ -> Na₂SO₃ + H₂O

=> m_Na2SO3 = 25,2g

Nhận thấy kết tủa lớn nhất khi có Mg(OH)₂ : 0,06 mol và Al(OH)₃ : 0,09 mol

n_OH^- = n_H⁺ + 2×n_Mg²⁺ + 3×n_Al³⁺ = 0,11 + 2×0,06 + 3×0,09 = 0,5 mol

→ 0,02V + 0,02V= 0,5 → V= 12,5 lít

Kết tủa nhỏ nhất khi chỉ có Mg(OH)₂ : 0,06 mol

n_OH^- = n_H⁺ + 2×n_Mg²⁺ + 4×n_Al³⁺ = 0,11 + 2×0,06 + 4×0,09 = 0,59 mol

→ 0,02 V + 0,02V = 0,59 → V= 14,75 lit

Đáp án C