Zn+2CH3COOH->(CH3COO)2Zn+H2

0,25-------0,5------------------------------0,25

n CH3COOH=\(\dfrac{30}{60}=0,5mol\)

=>m Zn=0,25.65=16,25g

=>VH2=0,25.22,4=5,6l

CH3COOH+C2H5OH->CH3COOC2H5+H2O

 1-------------------------------------1

n CH3COOH=1 mol

n C2H5OH=2,17 mol

=>C2H5OH dư

=>m CH3COOC2H5=1.88=88g

=>H=\(\dfrac{55}{88}100=62,5\%\)

1.\(n_{CH_3COOH}=\dfrac{30}{60}=0,5mol\)

\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

0,25          0,5                                           0,25 ( mol )

\(m_{Zn}=0,25.65=16,25g\)

\(V_{H_2}=0,25.22,4=5,6l\)

2.\(n_{CH_3COOH}=\dfrac{60}{60}=1mol\)

\(n_{C_2H_5OH}=\dfrac{100}{46}=2,17mol\)

\(n_{CH_3COOC_2H_5}=\dfrac{55}{88}=0,625mol\)

\(CH_3COOH+C_2H_5OH\rightarrow CH_3COOC_2H_5+H_2O\)

          1                 2,17                   0,625                        ( mol )

         0,625                                       0,625                      ( mol ) 

=> H tính théo CH3COOH

\(H=\dfrac{0,625}{1}.100=62,5\%\)

B học trường nào đấy

Câu 9:

\(n_{CH_3CHO}=\dfrac{5,28}{44}=0,12\left(mol\right)\)

PT: \(CH_3CHO+H_2\underrightarrow{t^o,Ni}CH_3CH_2OH\)

Theo PT: \(n_{H_2}=n_{CH_3CHO}=0,12\left(mol\right)\Rightarrow V_{H_2}=0,12.22,4=2,688\left(l\right)\)

Đáp án: A

Câu 12:

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

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

Theo PT: \(n_{C_2H_5OH}=2n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{C_2H_5OH}=0,3.46=13,8\left(g\right)\)

Đáp án: D

Giả sử KL có hóa trị n.

PT: \(2M+2nHCl\rightarrow2MCl_n+nH_2\)

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

Theo PT: \(n_M=\dfrac{2}{n}nH_2=\dfrac{0,4}{n}\left(mol\right)\)

\(\Rightarrow M_M=\dfrac{4,8}{\dfrac{0,4}{n}}=12n\left(g/mol\right)\)

Với n = 2 thì M_M = 24 (g/mol) là tm

Vậy: M là Mg.

Ta có: \(n_{MgCl_2}=n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

\(\Rightarrow m_{MgCl_2}=0,2.95=19\left(g\right)\)

nZn = 3,25/65=0,05 mol

2Zn + 2CH3COOH --> 2CH3COOZn + H2

0,05      0,05                   0,05               0,025            mol

=> VH2= 0,025*22,4=0,56 lít

mdd=(0,05*60*100)/20=15 g

b)mCH3COOZn = 0,05*124=6,2 g

Đặt \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Na}=b\left(mol\right)\end{matrix}\right.\)

PTHH:
\(Fe+2HCl\rightarrow FeCl_2+H_2\)

a--------------------------->a

\(2Na+2HCl\rightarrow2NaCl+H_2\)

b---------------------------->0,5b

Ta có: \(m_M=\dfrac{1}{2}.\left(m_{Fe}+m_{Na}\right)=\dfrac{1}{2}.\left(56a+23b\right)=28a+11,5b\left(g\right)\)

PTHH: \(M+2HCl\rightarrow MCl_2+H_2\)

        (a+0,5b)<----------------(a+0,5b)

\(\Rightarrow M_M=\dfrac{28a+11,5b}{a+0,5b}\\ \Rightarrow\dfrac{28a}{a}>M_M>\dfrac{11,5a}{0,5a}\\ \Leftrightarrow28>M_M>23\)

Vậy M là Magie (Mg)

Zn +2HCl->ZnCl2+H2

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

->VH2=0,1.22,4=2,24l

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