Gọi nFe = a (mol); nMg = b (mol)

56a + 24b = 8 (1)

nH2 = 4,48/22,4 = 0,2 (mol)

PTHH: 

Fe + 2HCl -> FeCl2 + H2

a ---> a ---> a ---> a

Mg + 2HCl -> MgCl2 + H2

b ---> b ---> b ---> b

a + b = 0,2 (2)

(1)(2) => a = b = 0,1 (mol)

mFe = 0,1 . 56 = 5,6 (g)

%mFe = 5,6/8 = 70%

%mMg = 100% - 70% = 30%

nHCl = 0,1 . 2 + 0,1 . 2 = 0,4 (mol)

CMddHCl = 0,4/0,1 = 4M

n_NaOH = 0,1.2 = 0,2 (mol)

PTHH: CH₃COOH + NaOH --> CH₃COONa + H₂O

                0,2<--------0,2

=> \(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,2}{0,15}=\dfrac{4}{3}M\)

nNaOH = 0,1 . 2 = 0,2 (mol)

PTHH: CH3COOH + NaOH -> CH3COONa + H2O

nCH3COOH = nNaOH = 0,2 (mol)

CM(CH3COOH) = 0,2/0,15 = 1,33M

$a\big)$

$n_{CH_3COOH}=\dfrac{100}{1000}.1=0,1(mol)$

$CH_3COOH+NaOH\to CH_3COONa+H_2O$

Theo PT: $n_{NaOH}=n_{CH_3COOH}=0,1(mol)$

$\to C\%_{NaOH}=\dfrac{0,1.40}{50}.100\%=80\%$

$b\big)$

$n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1(mol)$

$2CH_3COOH+Na_2CO_3\to 2CH_3COONa+CO_2+H_2O$

Theo PT: $\begin{cases} n_{CO_2}=n_{Na_2CO_3}=0,1(mol)\\ n_{CH_3COONa}=2n_{Na_2CO_3}=0,2(mol) \end{cases}$

$\to C\%_{CH_3COONa}=\dfrac{0,2.82}{60+10,6-0,1.44}.100\%\approx 24,77\%$

Chọn đáp án C

Chọn đáp án C

Chọn đáp án C

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

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

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

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

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

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

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

Đáp án B