a, \(n_{NaOH}=0,2.1=0,2\left(mol\right)\)

\(m_{NaOH}=0,2.40=8\left(g\right)\)

b, \(n_{H_2SO_4}=2.0,1=0,2\left(mol\right)\)

\(c,C\%=\dfrac{6}{200}.100\%=3\%\)

\(m_{NaCl}=\dfrac{200.8}{100}=16\left(g\right)\)

\(n_{Na}=\dfrac{2.3}{23}=0.1\left(mol\right)\)

\(n_{HCl}=0.4\cdot2=0.8\left(mol\right)\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

\(0.8..............0.8\)

\(2Na+2H_2O\rightarrow2NaOH+H_{_{ }2}\)

\(0.1........................0.1\)

\(n_{NaOH}=0.1< 0.8\)

Đề nhầm lãn !

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a) \(m_{HCl}=200\cdot7,3\%=14,6\left(g\right)\)

b) \(n_{NaOH}=0,5\cdot1=0,5\left(mol\right)\) \(\Rightarrow m_{NaOH}=0,5\cdot40=20\left(g\right)\)

c) \(n_{CuSO_4}=0,2\cdot1,5=0,3\left(mol\right)\) \(\Rightarrow m_{CuSO_4}=0,3\cdot160=48\left(g\right)\)

d) Bạn xem lại đề !

a) mHCl=200⋅7,3%=14,6(g)mHCl=200⋅7,3%=14,6(g)

b) nNaOH=0,5⋅1=0,5(mol)nNaOH=0,5⋅1=0,5(mol) ⇒mNaOH=0,5⋅40=20(g)⇒mNaOH=0,5⋅40=20(g)

c) nCuSO4=0,2⋅1,5=0,3(mol)nCuSO4=0,2⋅1,5=0,3(mol) ⇒mCuSO4=0,3⋅160=48(g)⇒mCuSO4=0,3⋅160=48(g)

d) Bạn xem lại đề !

\(n_{NaOH}=0.2\cdot0.5=0.1\left(mol\right)\)

\(n_{CuSO_4}=0.1\cdot2=0.2\left(mol\right)\)

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

\(0.1.............0.05...............0.05...........0.05\)

\(m_{Cu\left(OH\right)_2}=0.05\cdot98=4.9\left(g\right)\)

\(C_{M_{Na_2SO_4}}=\dfrac{0.05}{0.2+0.1}=0.167\left(M\right)\)

\(C_{M_{CuSO_4\left(dư\right)}}=\dfrac{0.2-0.05}{0.1}=1.5\left(M\right)\)

a) \(\left\{{}\begin{matrix}n_{CuSO_4}=0,3.1=0,3\left(mol\right)\\n_{BaCl_2}=0,1.2=0,2\left(mol\right)\end{matrix}\right.\)

PTHH:           \(CuSO_4+BaCl_2\rightarrow BaSO_4\downarrow+CuCl_2\)

Ban đầu:         0,3           0,2

Sau pư:           0,1            0              0,2             0,2

=> \(m_{kt}=m_{BaSO_4}=0,2.233=46,6\left(g\right)\)

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

                 0,1-------->0,2

                \(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)

                 0,2------>0,4

=> \(m_{ddNaOH}=\dfrac{\left(0,2+0,4\right).40}{15\%}=160\left(g\right)\)

\(n_{H_2SO_4}=0,15.1=0,15\left(mol\right)\)

PTHH: 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

Mol:      0,3           0,15

\(V_{ddKOH}=\dfrac{0,3}{1,5}=0,2\left(l\right)=200\left(ml\right)\)

  ⇒ Chọn B

Bài 1:

Ta có: \(n_{OH^-}=n_{Na^+}=n_{NaOH}=0,2.0,4=0,08\left(mol\right)\)

\(n_{H^+}=n_{Cl^-}=n_{HCl}=0,4.0,3=0,12\left(mol\right)\)

PT ion: \(OH^-+H^+\rightarrow H_2O\)

_____0,08_____0,12 (mol)

⇒ n_{OH^- (dư)} = 0,04 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\left[Na^+\right]=\frac{0,08}{0,6}\approx0,133M\\\left[Cl^-\right]=\frac{0,12}{0,6}=0,2M\\\left[OH^-\right]=\frac{0,04}{0,6}\approx0,066M\end{matrix}\right.\)

Câu 2:

Ta có: \(\Sigma n_{K^+}=n_{KCl}+2n_{K_2SO_4}=0,2.1,5+0,3.2.2=1,5\left(mol\right)\)

\(n_{Cl^-}=n_{KCl}=0,2.1,5=0,3\left(mol\right)\)

\(n_{SO_4^{2-}}=0,3.2=0,6\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\left[K^+\right]=\frac{1,5}{0,5}=3M\\\left[Cl^-\right]=\frac{0,3}{0,5}=0,6M\\\left[SO_4^{2-}\right]=\frac{0,6}{0,5}=1,2M\end{matrix}\right.\)

Bạn tham khảo nhé!