Để trung hòa 480 gam dung dịch axit axetic cần dùng 200 ml dung dịch NaOH 2M . Tính nồng độ % của axit axetic trong dung dịch ban đầu.
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\\ n_{NaOH}=0,2.0,1=0,02\left(mol\right)\\ n_{CH_3COOH}=n_{NaOH}=0,02\left(mol\right)\\ C_{MddCH_3COOH}=\dfrac{0,02}{0,1}=0,2\left(M\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2CH3COOH+Mg->(CH3COO)2Mg+H2
0,02---------------0,01-------0,01----------0,01
n muối=0,01mol
=>CM=\(\dfrac{0,02}{0,04}=0,5M\)
=>VH2=0,01.22,4=0,224l
CH3COOH+NaOH->CH3COONa+H2O
0,02--------------0,02
=>VNaOH=\(\dfrac{0,02}{0,75}=0,03l\)
a) \(n_{\left(CH_3COO\right)_2Mg}=\dfrac{1,42}{142}=0,01\left(mol\right)\)
PTHH: Mg + 2CH3COOH --> (CH3COO)2Mg + H2
0,01<-------0,02<------------0,01------->0,01
=> \(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,02}{0,04}=0,5M\)
b) VH2 = 0,01.22,4 = 0,224 (l)
c)
PTHH: NaOH + CH3COOH --> CH3COONa + H2O
0,02<------0,02
=> \(V_{dd.NaOH}=\dfrac{0,02}{0,75}=\dfrac{2}{75}\left(l\right)=\dfrac{80}{3}\left(ml\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(HCOOH+NaOH\rightarrow HCOONa+H_2O\)
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
\(n_{HCOOH}=a\left(mol\right),n_{CH_3COOH}=b\left(mol\right)\)
\(m_X=46a+60b=10.6\left(g\right)\left(1\right)\)
\(n_{NaOH}=a+b=0.2\left(mol\right)\left(2\right)\)
\(\Rightarrow a=b=0.1\)
\(m_{HCOOH}=0.1\cdot46=4.6\left(g\right)\)
\(m_{CH_3COOH}=6\left(g\right)\)
\(n_{H_2O}=0.2\left(mol\right)\)
\(BTKL:\)
\(m_{Muối}=10.6+0.2\cdot40-0.2\cdot18=15\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
$a\big)$
$n_{CH_3COOH}=\dfrac{100}{1000}.1=0,1(mol)$
$CH_3COOH+NaOH\to CH_3COONa+H_2O$
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: $\begin{cases} n_{CO_2}=n_{Na_2CO_3}=0,1(mol)\\ n_{CH_3COONa}=2n_{Na_2CO_3}=0,2(mol) \end{cases}$
$\to C\%_{CH_3COONa}=\dfrac{0,2.82}{60+10,6-0,1.44}.100\%\approx 24,77\%$
![](https://rs.olm.vn/images/avt/0.png?1311)
Chọn A
Gọi nồng độ mol ban đầu của HCl và H 2 S O 4 lần lượt là x và y (M)
![](https://rs.olm.vn/images/avt/0.png?1311)
CH3COOH + NaOH $\to$ CH3COONa + H2O
n NaOH = n CH3COOH = 50.6%/60 = 0,05(mol)
=> V dd NaOH = 0,05/2 = 0,025(lít) = 25(ml)
![](https://rs.olm.vn/images/avt/0.png?1311)
$a\big)$
$Zn+2CH_3COOH\to (CH_3COO)_2Zn+H_2$
$ZnO+2CH_3COOH\to (CH_2COO)_2Zn+H_2O$
Theo PT: $n_{Zn}=n_{H_2}=\frac{4,48}{22,4}=0,2(mol)$
$\to \%m_{Zn}=\frac{0,2.65}{21,1}.100\%\approx 61,61\%$
$\to \%m_{ZnO}=100-61,61=38,39\%$
$b\big)$
$n_{ZnO}=\frac{21,1-0,2.65}{81}=0,1(mol)$
Theo PT: $\sum n_{CH_3COOH}=2n_{Zn}+2n_{ZnO}=0,6(mol)$
$\to C_{M_{CH_3COOH}}=\dfrac{0,6}{\frac{200}{1000}}=3M$
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\)
\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
0,2 0,2 ( mol )
\(m_{Zn}=0,2.65=13g\)
\(\%m_{Zn}=\dfrac{13}{21,1}.100=61,61\%\)
\(\%m_{ZnO}=100\%-61,61\%=38,39\%\)
\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
0,2 0,4 ( mol )
\(n_{ZnO}=\dfrac{21,1-13}{81}=0,1mol\)
\(ZnO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2O\)
0,1 0,2 ( mol )
\(C_{M\left(CH_3COOH\right)}=\dfrac{0,4+0,2}{0,2}=3M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)
Ta có: \(n_{HCl}=0,2\cdot2=0,4\left(mol\right)=n_{NaOH}\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
Câu 2: Bạn xem lại đề !!
nNaOH = 0,2.2 = 0,4 (mol)
PTHH: NaOH + CH3COOH --> CH3COONa + H2O
0,4------>0,4
=> \(C\%_{ddCH_3COOH}=\dfrac{0,4.60}{480}.100\%=5\%\)