Giả sử có 100g dd CH3COOH 15%

Ta có:

nCH3COOH=\(\dfrac{100.15\%}{60}\)=0,25(mol)

PTHH:Ca(OH)2+2CH3COOH→(CH3COO)2Ca+2H2O

⇒nCa(OH)2=n(CH3COO)2Ca=0,125(mol)

⇒mdd(Ca(OH)2)=\(\dfrac{0,125.74}{X\%}\)=9,25

⇒mdd(spu)=\(\dfrac{9,25}{X\%+100}\)⇔C%=9,875%

\(\dfrac{0,125.158.100}{\dfrac{9,25}{X\%+100}}\)=9,875⇔x=9,25

$a\big)$

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

$CH_3COOH+NaOH\to CH_3COONa+H_2O$

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

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

* Ta có: \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

a. PTHH: \(CO_2+Ca\left(OH\right)_2--->CaCO_3\downarrow+H_2O\)

b. Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CO_2}=0,25\left(mol\right)\)

Đổi 100ml = 0,1 lít

=> \(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5M\)

* PTHH: X₂O₃ + 3H₂SO₄ ---> X₂(SO₄)₃ + 3H₂O

Đổi 600ml = 0,6 lít

Ta có: \(n_{H_2SO_4}=1.0,6=0,6\left(mol\right)\)

Theo PT: \(n_{X_2O_3}=\dfrac{1}{3}.n_{H_2SO_4}=\dfrac{1}{3}.0,6=0,2\left(mol\right)\)

=> \(M_{X_2O_3}=\dfrac{32}{0,2}=160\left(g\right)\)

Ta có: \(M_{X_2O_3}=NTK_X.2+16.3=160\left(g\right)\)

=> NTK_X = 56(đvC)

Vậy X là sắt (Fe)

=> CTHH là Fe₂O₃

a) $CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$

b) $n_{Ca(OH)_2} = n_{CO_2} = \dfrac{5,6}{22,4} = 0,25(mol)$

$\Rightarrow C_{M_{Ca(OH)_2}} = \dfrac{0,25}{0,1} = 2,5M$

c) $n_{CaCO_3} = n_{CO_2} = 0,25(mol)$

$\Rightarrow m_{CaCO_3} = 0,25.100 = 25(gam)$

\(m_{CH_3COOH}=12\%.100=12\left(g\right)\\ n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\)

PTHH: CH₃COOH + NaOH ---> CH₃COONa + H₂O

               0,2--------->0,2------------>0,2

\(m_{NaOH}=0,2.40=8\left(g\right)\\ m_{ddNaOH}=\dfrac{8}{8,4\%}=\dfrac{2000}{21}\left(g\right)\\ m_{ddCH_3COONa}=\dfrac{2000}{21}+100=\dfrac{4100}{21}\left(g\right)\\ m_{CH_3COONa}=0,2.82=16,4\left(g\right)\\ C\%_{CH_3COONa}=\dfrac{16,4}{\dfrac{4100}{21}}.100\%=8,4\%\)

`=>` Gợi ý:

`CH3COOH + NaHCO3 => CH3COONa + CO2 + H2O`

`m<sub>CH3COOH</sub> = 100x12/100 = 12` (g)

`==> n<sub>CH3COOH</sub> = m/M = 12/60 = 0.2` (mol)

Theo pt: `=> n<sub>NaHCO3</sub> = 0.2` (mol)

`==> m<sub>NaHCO3</sub> = n.M = 0.2x84 =16.8` (g)

`==> m<sub>dd NaHCO3</sub> = 16.8x100/8.4 = 200` (g)

Ta có: `n<sub>CH3COONa</sub> = 0.2` (mol)

\(a.n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ a.Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ 0,25.......0,25............0,25..........0,25\left(mol\right)\\ C_{MddCa\left(OH\right)_2}=\dfrac{0,25}{0,1}=2,5\left(M\right)\\ b.m_{\downarrow}=m_{CaCO_3}=100.0,25=25\left(g\right)\)

Coi 

\(m_{dd\ NaOH} = 100\ gam\\ \Rightarrow n_{NaOH} = \dfrac{100.10\%}{40} = 0,25(mol)\)

CH₃COOH + NaOH → CH₃COONa + H₂O

0,25................0,25.................0,25......................(mol)

\(m_{CH_3COONa} = 0,25.82 = 20,5(gam)\\ \Rightarrow m_{dd\ sau\ pư} = \dfrac{20,5}{10,25\%} = 200(gam)\\ \Rightarrow m_{dd\ axit\ axetic} = 200 -100 = 100(gam)\)

Vậy :

\(C\%_{CH_3COOH} = \dfrac{0,25.60}{100}.100\% = 15\%\)

\(n_{HCl}=0,1.0,2=0,02\left(mol\right)\)

Pt : \(2HCl+Ca\left(OH\right)_2\rightarrow CaCl_2+2H_2O\)

       0,02---->0,01---------->0,01

a) Nồng độ mol đề cho rồi mà nhỉ 

b) \(m_{muôi}=m_{CaCl2}=0,01.111=1,11\left(g\right)\)

a, \(n_{KOH}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

PT: \(CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\)

Theo PT: \(n_{CH_3COOH}=n_{KOH}=0,15\left(mol\right)\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\)

b, \(n_{Na_2CO_3}=0,2.0,5=0,1\left(mol\right)\)

PT: \(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)

Xét tỉ lệ: \(\dfrac{0,15}{2}< \dfrac{0,1}{1}\), ta được Na₂CO₃ dư.

Theo PT: \(n_{CO_2}=\dfrac{1}{2}n_{CH_3COOH}=0,075\left(mol\right)\Rightarrow V_{CO_2}=0,075.22,4=1,68\left(l\right)\)