300ml = 0,3l

\(n_{HNO3}=1.0,3=0,3\left(mol\right)\)

Pt : \(NaOH+HNO_3\rightarrow NaNO_3+H_2O|\)

          1               1               1              1

         0,3            0,3            0,3

\(n_{NaOH}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

200ml = 0,2l

\(C_{M_{NaOH}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

\(n_{NaNO3}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

⇒ \(m_{NaNO3}=0,3.85=25,5\left(g\right)\)

Sau phản ứng :

\(V_{dd}=0,2+0,3=0,5\left(l\right)\)

\(C_{M_{NaNO3}}=\dfrac{0,3}{0,5}=0,6\left(M\right)\)

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\(n_{HNO_3}=0,3\left(mol\right)\)

\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)

Theo PT: \(n_{NaOH}=n_{NaNO_3}=n_{HNO_3}=0,3\left(mol\right)\)

\(\Rightarrow CM_{NaOH}=\dfrac{0,3}{0,2}=1,5M\)

\(m_{NaNO_3}=0,3.85=25,5\left(g\right)\)

$n_{NaOH} = \dfrac{50.10\%}{40} = 0,125(mol)$
$CH_3COOH + NaOH \to CH_3COONa + H_2O$
Theo PTHH : 

$n_{CH_3COOH} = n_{CH_3COONa} = n_{NaOH} = 0,125(mol)$
$m_{dd\ CH_3COOH} = \dfrac{0,125.60}{8\%} = 93,75(gam)$
$m_{dd\ sau\ pư} = m_{dd\ CH_3COOH} + m_{dd\ NaOH} = 143,75(gam)$
$C\%_{CH_3COONa} = \dfrac{0,125.82}{143,75}.100\% = 7,13\%$

\(n_{H_2SO_4}=0,4\cdot1=0,4mol\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)

 0,8              0,4

\(V_{NaOH}=\dfrac{0,8}{0,5}=1,6l\)

\(n_{H_2SO_4}=0.4\cdot1=0.4\left(mol\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

\(0.8.............0.4..............0.4\)

\(V_{dd_{NaOH}}=\dfrac{0.8}{0.5}=1.6\left(l\right)\)

\(V_{dd}=0.4+1.6=2\left(l\right)\)

\(C_{M_{Na_2SO_4}}=\dfrac{0.4}{2}=0.2\left(M\right)\)

Câu 8 : 

\(n_{MgCO3}=\dfrac{42}{84}=0,5\left(mol\right)\)

Pt : \(2CH_3COOH+MgCO_3\rightarrow\left(CH_3COO\right)_2Mg+CO_2+H_2O\)

                 1                  0,5                                         0,5

a) \(V_{CO2\left(dktc\right)}=0,5.22,4=11,2\left(l\right)\)

b) \(V_{CH3COOH}=\dfrac{1}{2}=0,5\left(l\right)\)

c) Pt : \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

                  1                                         1

300ml = 0,3l

\(C_{MCH3COONa}=\dfrac{1}{0,3}=\dfrac{10}{3}\left(M\right)\)

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Đổi 300ml = 0,3 lít

Ta có: \(n_{H_2SO_4}=0,3.0,5=0,15\left(mol\right)\)

PTHH: H₂SO₄ + 2KOH ---> K₂SO₄ + 2H₂O

a. Theo PT: \(n_{KOH}=2.n_{H_2SO_4}=2.0,15=0,3\left(mol\right)\)

\(\Rightarrow V_{dd_{KOH}}=\dfrac{0,3}{0,2}=1,5\left(lít\right)\)

b. Theo PT: \(n_{K_2SO_4}=n_{H_2SO_4}=0,15\left(mol\right)\)

\(\Rightarrow m_{K_2SO_4}=0,15.174=26,1\left(g\right)\)

c. Ta có: \(V_{dd_{K_2SO_4}}=V_{dd_{H_2SO_4}}=0,3\left(lít\right)\)

\(\Rightarrow C_{M_{K_2SO_4}}=\dfrac{0,15}{0,3}=0,5M\)

\(n_{NaOH}=0.5\cdot0.5=0.25\left(mol\right)\)

\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)

\(0.25.........0.125...........0.125\)

\(C_{M_{Na_2CO_3}}=\dfrac{0.125}{0.5}=0.25\left(M\right)\)

100ml = 0,1l

\(n_{H2SO4}=3.0,1=0,3\left(mol\right)\)

a) Pt : \(H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O|\)

                1             2                1              2

              0,3           0,6             0,3

b) \(n_{K2SO4}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

⇒ \(m_{K2SO4}=0,3.174=52,2\left(g\right)\)

c) \(n_{KOH}=\dfrac{0,3.2}{1}=0,6\left(mol\right)\)

\(V_{ddKOH}=\dfrac{0,6}{2}=0,3\left(l\right)\)

d) \(V_{ddspu}=0,1+0,3=0,4\left(l\right)\)

\(C_{M_{K2SO4}}=\dfrac{0,3}{0,4}=0,75\left(M\right)\)

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