\(n_{NaOH}=1,5\left(mol\right)\)

\(V=\dfrac{1,5}{5}=0,3\left(l\right)=300\left(ml\right)\)

\(n_{NaOH\left(ct\right)}=\dfrac{m_{NaOH}}{M_{NaOH}}=1,5\left(mol\right)\)

\(V_{ddNaOH}=\dfrac{n_{ctNaOH}}{C_{M_{NaOH}}}=0,3\left(l\right)\)

\(n_{NaOH}=\frac{60}{40}=1,5(mol)\\ V_{ddNaOH}=\frac{1,5}{5}=0,3(l)=300(ml)\)

Chọn B

a)

$n_{Al_2O_3} = \dfrac{5,1}{102} = 0,05(mol)$
$Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O$
$n_{HCl} = 6n_{Al_2O_3} = 0,3(mol)$
$\Rightarrow V = \dfrac{0,3}{4} = 0,075(lít)$

b)

$Al_2O_3 + 2NaOH \to 2NaAlO_2 + 2H_2O$
$n_{NaOH} = 2n_{Al_2O_3} = 0,1(mol)$
$V_{dd\ NaOH} = \dfrac{0,1}{10} = 0,01(lít)$

a) \(n_{CH_3COOH}=0,1.0,3=0,03\left(mol\right)\)

PTHH: CH₃COOH + NaOH --> CH₃COONa + H₂O

                  0,03---->0,03--------->0,03

=> \(V_{dd.NaOH}=\dfrac{0,03}{1,5}=0,02\left(l\right)\)

b) m_CH3COONa = 0,03.82 = 2,46 (g)

c) \(C_{M\left(CH_3COONa\right)}=\dfrac{0,03}{0,1+0,02}=0,25M\)

\(n_{CuO}=\dfrac{8}{80}=0.1\left(mol\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

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

\(n_{NaOH}=0.24\cdot0.5=0.12\left(mol\right)\)

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

\(0.12..........0.06\)

\(n_{H_2SO_4}=0.1+0.06=0.16\left(mol\right)\)

\(V_{dd_{H_2SO_4}}=\dfrac{0.16}{1}=0.16\left(l\right)\)

\(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0.06}{0.16}=0.375\left(M\right)\)

\(C_{M_{CuSO_4}}=\dfrac{0.1}{0.16}=0.625\left(M\right)\)

\(n_{Na}=\dfrac{13,8}{23}=0,6\left(mol\right)\\ pthh:Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\) 
          0,6                    0,6                  0,3 
\(V_{H_2}=0,3.22,4=6,72\left(l\right)\\ c,m_{\text{dd}}=13,8+286,8-\left(0,3.2\right)=300\left(g\right)\\ C\%=\dfrac{0,6.40}{300}.100\%=8\%\)

\(n_{Na}\) = \(\dfrac{13,8}{23}\) = 0,6 mol

Theo PTHH:

a) \(2Na+2H_2O\underrightarrow{t^o}2NaOH+H_2\)

     2          2           2                1 (mol)

    0,6 \(\rightarrow\) 0,6   \(\rightarrow\) 0,6     \(\rightarrow\)    0,3 (mol)

b) \(V_{H_2}\) = 0,3.22,4 = 6,72l

c) \(m_{dd}\) = 13,8 + 286,8  - 0,3.2 = 300g

\(C\%\) = \(\dfrac{0,6.40}{300}\).100% = 8%

Gọi lượng NaOH cần hòa tan vào dd A là x (g)

Ta có công thức :

Vậy phải trộn thêm 6,52g NaOH vào dung dịch A để được dung dịch A để được dung dịch NaOH 8%.