PTHH: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

Ta có: \(n_{NaOH}=\dfrac{30\cdot20\%}{40}=0,15\left(mol\right)=n_{CH_3COONa}=n_{CH_3COOH}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CH_3COONa}=0,15\cdot82=12,3\left(g\right)\\C_{M_{CH_3COOH}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\end{matrix}\right.\)

\(CH_3COOH+NaCl\rightarrow CH_3COONa+HCl\)

\(2CH_3COOH+CaCO_3\rightarrow\left(CH_3COO\right)_2Ca+H_2O+CO_2\)

   2                         1                     1                      1         1    (mol)

 0,08                    0,04                0,04              0,04         0,04 (mol)

\(nCO_2=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

\(mCaCO_3=0,04.100=4\left(g\right)\)

=> \(mNaCl=12,5-4=8,5\left(g\right)\)

( không thấy hh B ) 

c ) .

\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

 1             2               1                  1     (mol)

0,04       0,08           0,04        0,04 (mol)

\(mNa_2CO_3=0,04.106=4,24\left(g\right)\)

\(mNa_2CO_{3\left(thựctế\right)}=\)\(\dfrac{4,24.85\%}{100\%}=3,604\left(g\right)\)

PTHH: CH₃COOH + NaHCO₃ ➞ CH₃COONa + H₂O + CO₂

Ta có: mNAHCO3=(200.84)/100= 16.8 gam

nNAHCO3= 16.8/84= 0.2 mol

mCH3COOH= 0.2*60= 12 gam

Câu a) mddCH3COOH= (12*100)/6= 200 gam

Câu b) mddCH3COONA= mddCH3COOH + mdd NAHCO3= 200+200=400 gam

mCH3COONA= 0.2*82=16.4 gam

C%dd CH3COONA= (16.4*100)/400= 4.1%

Chúc bạn học tốt

Bài 1

\(a,n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ CuO+2HCl\xrightarrow[]{}CuCl_2+H_2O\\ n_{CuCl_2}=n_{CuO}=0,2mol\\ m_{CuCl_2}=0,2.135=27\left(g\right)\\ b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ C_{MHCl}=\dfrac{0,4}{0,5}=0,8\left(M\right)\)

Bài 5

\(a,n_{NaOH}=0,2.1=0,2\left(mol\right)\\ 2NaOH+H_2SO_4\xrightarrow[]{}Na_2SO_4+2H_2O\\ n_{H_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MH_2SO_4}=\dfrac{0,1}{0,4}=0,25\left(M\right)\\ b,n_{Na_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MNa_2SO_4}=\dfrac{0,1}{0,2+0,4}=\dfrac{1}{6}\left(M\right)\\ c,m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)

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 :

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

Gọi n_Na2CO3 = x (mol)

Na₂CO₃ + 2HCl \(\rightarrow\) 2NaCl + H₂O + CO₂

  x         \(\rightarrow\)    2x   \(\rightarrow\)    2 x                                    (mol)

C%_(NaCl) = \(\frac{2.58,5x}{200+120}\) . 100% = 20%

=> x =0,547 (mol)

m_Na2CO3 = 0,547 . 106 = 57,982 (g)

m_HCl = 2 . 0,547 . 36,5 =39,931 (g)

C%_(Na2CO3) =\(\frac{57,892}{200}\) . 100% = 28,946%

C%(HCl) = \(\frac{39,931}{120}\) . 100% = 33,28%

a) \(n_{CH_3COOH}=\dfrac{120.20\%}{60}=0,4\left(mol\right)\)

\(n_{Na_2CO_3}=\dfrac{53.30\%}{106}=0,15\left(mol\right)\)

PTHH: Na₂CO₃ + 2CH₃COOH --> 2CH₃COONa + CO₂ + H₂O

Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,4}{2}\) => Na₂CO₃ hết, CH₃COOH dư

PTHH: Na₂CO₃ + 2CH₃COOH --> 2CH₃COONa + CO₂ + H₂O

                0,15-------->0,3-------------->0,3------->0,15

=> \(m_{CH_3COONa}=0,3.82=24,6\left(g\right)\)

b) m_{dd sau pư} = 120 + 53 - 0,15.44 = 166,4 (g)

=> \(C\%=\dfrac{24,6}{166,4}.100\%=14,78\text{%}\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

           0,2<--0,4<------0,2<-----0,2

=> \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,2.65}{21,1}.100\%=61,61\%\\\%m_{ZnO}=100\%-61,61\%=38,39\%\end{matrix}\right.\)

\(n_{ZnO}=\dfrac{21,1-0,2.65}{81}=0,1\left(mol\right)\)

PTHH: ZnO + 2HCl ---> ZnCl₂ + H₂O

            0,1---->0,2------>0,1

=> \(C\%_{HCl}=\dfrac{\left(0,2+0,4\right).36,5}{200}.100\%=10,95\%\)

\(m_{mu\text{ố}i}=m_{ZnCl_2}=\left(0,1+0,2\right).136=40,8\left(g\right)\)