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) V_H2 = 0,01.22,4 = 0,224 (l)

c) 

PTHH: NaOH + CH₃COOH --> CH₃COONa + H₂O

             0,02<------0,02

=> \(V_{dd.NaOH}=\dfrac{0,02}{0,75}=\dfrac{2}{75}\left(l\right)=\dfrac{80}{3}\left(ml\right)\)

\(a.2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\\ n_{\left(CH_3COO\right)_2Mg}=0,01\left(mol\right)\\ n_{CH_3COOH}=2n_{\left(CH_3COO\right)_2Mg}=0,02\left(mol\right)\\ \Rightarrow CM_{CH_3COOH}=\dfrac{0,02}{0,04}=0,5M\\ b.n_{H_2}=n_{\left(CH_3COO\right)_2Mg}=0,01\left(mol\right)\\ \Rightarrow V_{H_2}=0,01.22,4=0,224\left(l\right)\\ c.CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\\ n_{CH_3COOH}=n_{NaOH}=0,02\left(mol\right)\\ \Rightarrow V_{NaOH}=\dfrac{0,02}{0,75}=0,02667\left(l\right)=26,67ml\)

\(n_{HCl}=0,2.0,2=0,04\left(mol\right)\)

Để trung hòa thì:

\(n_{NaOH}=n_{HCl}\)

\(\Leftrightarrow0,1.V_{NaOH}=0,04\)

\(\Leftrightarrow V_{NaOH}=0,4\left(l\right)=400\left(ml\right)\)

PTHH: \(HCl+NaOH\rightarrow NaCl+H_2O\)

\(\Rightarrow n_{NaCl}=n_{HCl}=0,04\left(mol\right)\)

\(V_{dd}=0,2+0,4=0,6\left(l\right)\)

\(\Rightarrow C_{M\left(NaCl\right)}=\dfrac{0,04}{0,6}=0,67M\)

Chọn A

Gọi nồng độ mol ban đầu của HCl và H 2 S O 4 lần lượt là x và y (M)

Đặt x, y là số mol HCl và  H 2 SO 4  trong 40 ml dung dịch A.

HCl + NaOH → NaCl +  H 2 O

H 2 SO 4 + 2NaOH →  Na 2 SO 4  + 2 H 2 O

Số mol NaOH: x + 2y = 1.60/1000 = 0,06 mol (1)

Khối lượng 2 muối : 58,5x + 142y = 3,76 (2)

Từ (1), (2), giải ra : x = 0,04 ; y = 0,01.

C M HCl  = 0,04/0,04 = 1(mol/l)

C M H 2 SO 4 = 0,01/0,04 = 0,25 (mol/l)

200ml = 0,2l

\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

        1         2              1           1

     0,3      0,6             0,3        0,3

a) \(n_{ZnCl2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

\(C_{M_{ZnCl2}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

b) \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)

c) Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)

              1            1              1            1

            0,6          0,6

\(n_{NaOH}=\dfrac{0,6.1}{1}=0,6\left(mol\right)\)

\(m_{NaOH}=0,6.40=24\left(g\right)\)
\(m_{ddNaOH}=\dfrac{24.100}{20}=120\left(g\right)\)

 Chúc bạn học tốt

a.250ml=0,25l ; nHCl=0,25.1,5=0,375mol

KOH+HCl->KCl+H2O

1mol 1mol 1mol

0,375 0,375 0,375

VKOh=0,375/2=0,1875l

b.CM KCL=0,375/0,25=1,5M

c.NaOH+HCL=NaCl+H2O

1mol 1mol

0,375 0,375

mdd NaOH=0,375.40.100/10=150g

n_HCl = C_M . V = 1,5 . 0,25 = 0,375 mol

PTHH: KOH + HCl -------> KCl + H₂O

Pt: 1 1 1 1 (mol)

Pư 0,375 <-0,375 -----> 0,375-> 0,375 (mol)

a) V_KOH = \(\dfrac{n}{C_M}\)= 0,1875 l

b) C_M KCl = \(\dfrac{n}{V}\)= \(\dfrac{0,375}{0,25+0,1875}\)\(\approx\)0,86 M

c) PTHH: NaOH + HCL ----> NaCl + H₂O

Pt: 1 1 1 1 (mol)

Pư 0,375 <- 0,375 ------>0,375 --> 0,375 (mol)

m_ctNaOH = n . M = 15 g

m_{ddNaOH = \(\dfrac{15.100\%}{10\%}\)=} 150 g

:3

.

\(n_{HCl}=0,2.2=0,04\left(mol\right)\)

Pt: \(NaOH+HCl\rightarrow NaCl+H_2O\)

0,04mol \(\leftarrow\) 0,04mol \(\rightarrow\) 0,04mol

\(V_{NaOH}=\dfrac{0,04}{0,1}=0,4\left(l\right)\)

\(\Sigma_{V\left(spu\right)}=0,2+0,4=0,6\left(l\right)\)

\(C_{M_{NaCl}}=\dfrac{0,04}{0,6}=0,067M\)

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

0,02mol \(\leftarrow\) 0,04mol\(\rightarrow\) 0,02mol

\(m_{Ca\left(OH\right)_2}=0,02.74=1,48\left(g\right)\)

\(m_{dd}=\dfrac{1,48.100}{5}=29,6\left(g\right)\)

\(C\%_{CaCl_2}=\dfrac{0,02.111.100}{29,6}=7,5\%\)

1.NaOH+HCl--->NaCl+H2O

nNaOH=(200.10%)/40=0,5
=>nHCl=nNaOH=0,5
=>mddHCl=(0,5.36,5)/3,65%=500 g

2:a,2NaOH+H2SO4−−>Na2SO4+H2O2
Theo pthh, ta có: nNaOH=2.nH2SO4=0,4mol
-->mNaOH=16g
-->md/dNaOH=80g
b, Ta có: nKOH=0,4mol
-->md/dKOH=400g
-->V=383ml

1.NaOH + HCl ---> NaCl + H2O
0,5--------0,5
mctNaOH=mdd.C%/100=2000/100=20g
=> nNaOH=20/40=0,5mol
=>mctHCl=0,5.36,5=18,25
=>mddHCl=mct.100/C%=18,25.100/3,65=500...