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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: \(n_{CO_2}=\dfrac{1}{2}n_{CH_3COOH}=0,075\left(mol\right)\Rightarrow V_{CO_2}=0,075.22,4=1,68\left(l\right)\)
CH3COOH + NaOH => CH3COONa + H2O
500 ml dd = 0.5 l dd
mNaOH = 20x30/100 = 6 (g)
nNaOH = m/M = 6/40 = 0.15 (mol)
Theo phương trình => nCH3COOH = 0.15 (mol)
CM dd CH3COOH = n/V = 0.15/0.5 = 0.3M
2CH3COOH + Na2CO3 => 2CH3COONa + CO2 + H2O
nNa2CO3 = 0.5x0.2 = 0.1 (mol)
Theo phương trình ==> nCO2 = 0.1 (mol)
VCO2 = n x 22.4 = 0.1 x 22.4 = 2.24 (l)
nNaOH = 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.b.\(n_{Mg}=\dfrac{4,8}{24}=0,2mol\)
\(2Mg+2CH_3COOH\rightarrow2\left(CH_3COO\right)_2Mg+H_2\)
0,2 0,2 0,2 ( mol )
\(C_{M_{CH_3COOH}}=\dfrac{0,2}{0,2}=1M\)
\(m_{\left(CH_3COO\right)_2Mg}=0,2.142=28,4g\)
c.Sửa đề: thu được 9,2g este
\(n_{CH_3COOC_2H_5}=\dfrac{9,2}{88}=0,1mol\)
\(CH_3COOH+C_2H_5OH\rightarrow CH_3COOC_2H_5+H_2O\)
Thực tế: 0,2 0,1 ( mol )
Lý thuyết: 0,1 0,1 ( mol )
\(H=\dfrac{0,1}{0,2}.100=50\%\)
$a\big)$
$n_{CH_3COOH}=\dfrac{100}{1000}.1=0,1(mol)$
$CH_3COOH+NaOH\to CH_3COONa+H_2O$
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: $\begin{cases} n_{CO_2}=n_{Na_2CO_3}=0,1(mol)\\ n_{CH_3COONa}=2n_{Na_2CO_3}=0,2(mol) \end{cases}$
$\to C\%_{CH_3COONa}=\dfrac{0,2.82}{60+10,6-0,1.44}.100\%\approx 24,77\%$
Số mol của Mg là 4,8:24=0,2 (mol).
Mg (0,2 mol) + 2CH3COOH (0,4 mol) \(\rightarrow\) Mg(OOCCH3)2 (0,2 mol) + H2.
a) Nồng độ mol cần tìm là 0,4:0,2=2 (mol/l).
b) Khối lượng của magie axetat là 0,2.142=28,4 (g).
Bạn kiểm tra lại giúp mình câu c nhé!
\(m_{CH_3COOH}=12\%.100=12\left(g\right)\\ n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\)
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`
`mCH3COOH = 100x12/100 = 12` (g)
`==> nCH3COOH = m/M = 12/60 = 0.2` (mol)
Theo pt: `=> nNaHCO3 = 0.2` (mol)
`==> mNaHCO3 = n.M = 0.2x84 =16.8` (g)
`==> mdd NaHCO3 = 16.8x100/8.4 = 200` (g)
Ta có: `nCH3COONa = 0.2` (mol)
a, \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PT: \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
Theo PT: \(n_{CH_3COOH}=2n_{Mg}=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
b, \(n_{\left(CH_3COO\right)_2Mg}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow m_{\left(CH_3COO\right)_2Mg}=0,2.142=28,4\left(g\right)\)
a) nNaOH = \(\dfrac{40.20}{100.40}\)= 0,2 mol ,
pt: CH3COOH + NaOH \(\rightarrow\) CH3COONa +H2O
Theo pt nCH\(_3\)COOH = n NaOH = 0,2 mol
500 ml = 0,5 l
CM (CH\(_3\)COOH) = \(\dfrac{0,2}{0,5}\)= 0,4 M
b) nNa\(_2\)CO\(_3\)= 2.0,15 = 0,3 mol
pt: Na\(_2\)CO\(_3\) + 2CH\(_3\)COOH \(\rightarrow\) 2CH\(_3\)COONa + H2O + CO2
Tỉ lệ : nNa\(_2\)CO\(_3\) : n CH\(_3\)COOH = \(\dfrac{0,3}{1}\):\(\dfrac{0,2}{2}\)= 0,3 : 0,1
0,3>0,1 ,nên trong p ứ Na\(_2\)CO\(_3\) dư
CO2 sẽ tính theo CH\(_3\)COOH
nCO\(_2\)= 0,1 mol \(\rightarrow\) VCO\(_2\) = 22,4.0,1 = 2,24 l
a) nNaOH = 40.20100.4040.20100.40= 0,2 mol ,
pt: CH3COOH + NaOH →→ CH3COONa +H2O
Theo pt nCH33COOH = n NaOH = 0,2 mol
500 ml = 0,5 l
CM (CH33COOH) = 0,20,50,20,5= 0,4 M
b) nNa22CO33= 2.0,15 = 0,3 mol
pt: Na22CO33 + 2CH33COOH →→ 2CH33COONa + H2O + CO2
Tỉ lệ : nNa22CO33 : n CH33COOH = 0,310,31:0,220,22= 0,3 : 0,1
0,3>0,1 ,nên trong p ứ Na22CO33 dư
CO2 sẽ tính theo CH33COOH
nCO22= 0,1 mol →→ VCO22 = 22,4.0,1 = 2,24 l