a, 2CH3COOH + Na2CO3 ---> 2CH3OONa + H20 + CO2 

C2H5OH không tác dụng

b, đặt C2H5OH =x , CH3COOH=y 

Ta có hệ 46x + 60y = 21,4 và y=\(\frac{10}{M_{CaCO3}}=\frac{10}{100}=0,1\)

=> x và y => %

\(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

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

0,04<------------0,04

\(CH_3COOH+Na\rightarrow CH_3COONa+\dfrac{1}{2}H_2\uparrow\)

0,04----------------------------------------->0,02

\(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\uparrow\)

0,005<--------------------------------0,01

\(\rightarrow m_{C_2H_5OH}=0,005.46=0,23\left(g\right)\)

2C2H5OH + 2Na--> 2C2H5Na + H2

  a                                                 a/2           mol

2CH3COOH + 2Na --> 2CH3COONa + H2

 b                                                             b/2             mol

n khí = 3,36/22,4=0,15 mol

=> a/2 + b/2 =0,15

và 46a + 60 b =15,2 

=> a=0,2 mol  : b=0,1 mol

=> mC2H5OH = 0,2 * 46=9,2 g

=>% mC2H5OH = 9,2*100/15,2=60,53%

% mCH3COOH = 100 - 60 ,53=39,47 %

n H2 =0,3 mol

n NaOH =0,2 mol

=>n OH= n H+ =0,2 mol

=>n CH3COOH=0,2 mol

=>m CH3COOH=0,2.60=12g

->n C2H5OH=0,1 mol

=>m C2H5OH=4,6g

=>mC2H5OH=\(\dfrac{4,6}{4,6+12}100=27,71\%\)

=>maxit=72,29%

b)

CH3COOH+C2H5OH->CH3COOC2H5+H2O

                           0,1-------------0,1

=>H=82%

=>m CH3COOC2H5=0,082.88=7,216g

a, \(n_{C_2H_4}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)

PT: \(C_2H_4+H_2O\underrightarrow{^{t^o,xt}}C_2H_5OH\)

Theo PT: \(n_{C_2H_5OH\left(LT\right)}=n_{C_2H_4}=0,7\left(mol\right)\)

Mà: H = 90%

\(\Rightarrow n_{C_2H_5OH\left(TT\right)}=0,7.90\%=0,63\left(mol\right)\)

\(\Rightarrow m_{C_2H_5OH\left(TT\right)}=0,63.46=28,98\left(g\right)\)

\(\Rightarrow V_{C_2H_5OH}=\dfrac{28,98}{0,8}=36,225\left(ml\right)\)

b, \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)

Theo PT: \(n_{CH_3COOH}=n_{C_2H_5OH}=0,63\left(mol\right)\)

\(\Rightarrow m_{CH_3COOH}=0,63.60=37,8\left(g\right)\)

\(\Rightarrow m_{ddCH_3COOH}=\dfrac{37,8}{5\%}=756\left(g\right)\)

- Đặt \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\Rightarrow27a+24b=10,2\left(1\right)\)

Khí thu được sau p/ứ là khí H₂: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

2                                         3       (mol)

a                                        3/2 a   (mol)

\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

1                                        1   (mol)

b                                         b   (mol)

Từ hai PTHH trên ta có: \(\dfrac{3}{2}a+b=0,5\left(2\right)\)

\(\left(1\right),\left(2\right)\) ta có hệ: \(\left\{{}\begin{matrix}27a+24b=10,2\\\dfrac{3}{2}a+b=0,5\end{matrix}\right.\)

Giải ra ta có \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)

a) \(\%Al=\dfrac{m_{Al}}{m_{hh}}.100\%=\dfrac{0,2.27}{10,2}.100\%\approx52,94\%\)

\(\%Mg=100\%-\%Al=100\%-52,94=47,06\%\)

b)

 \(3H_2+Fe_2O_3\rightarrow^{t^0}2Fe+3H_2O\)

3               1              2  (mol)

0,5            1/6         1/3  (mol)

\(m_{Fe}=\dfrac{1}{3}.56=\dfrac{56}{3}\left(g\right)\)

\(m_{Fe_2O_3\left(pứ\right)}=\dfrac{1}{6}.160=\dfrac{80}{3}\left(g\right)\)

\(m_{Fe_2O_3\left(dư\right)}=60-m_{Fe}=60-\dfrac{56}{3}=\dfrac{124}{3}\left(g\right)\)

\(a=\dfrac{124}{3}+\dfrac{80}{3}=68\left(g\right)\)