D

D

D

\(n_{Mg}=\dfrac{m_{Mg}}{M_{Mg}}=\dfrac{4.8}{24}=0,2mol\) 

đổi 200 ml = 0,2 l

PTHH: Mg + 2HCl \(\rightarrow\) MgCl₂ + H₂

TL;        1        2           1          1

mol:     0,2 \(\rightarrow\) 0,2               \(\rightarrow\) 0,2

b. \(C_{M_{ddHCl}}=\dfrac{n_{HCl}}{V_{dd_{HCl}}}=\dfrac{0,2}{0,2}=1M\)  

\(c.V_{H_2}=n_{H_2}.22,4=0,2.22,4=4,48l\) 

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

a) PTHH : \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b) \(n_{HCl}=2n_{Mg}=0,4\left(mol\right)\Rightarrow C_{MddHCl}=\dfrac{0,4}{0,2}=2M\)

c) \(n_{Mg}=n_{H2}=0,2\left(mol\right)\Rightarrow V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

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

\(n_{\left(CH_3COO\right)_2Zn}=\dfrac{14,2}{183}\approx0.077mol\)

Theo pt: nH₂ = n(CH₃COO)₂Zn = 0,077mol

=> VH₂ = 1,7248l

b) Theo pt: nCH₃COOH = 2n(CH₃COO)₂Zn = 0,154 mol

=> CM_CH3COOH = 0,154 : 0,25 = 0,616M

\(n_{HCl}=0,3.2=0,6\left(mol\right)\\a, 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b,n_{H_2}=\dfrac{3}{6}.0,6=0,3\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ c,n_{Al}=n_{AlCl_3}=\dfrac{2}{6}.0,6=0,2\left(mol\right)\\ m_{Al}=0,2.27=5,4\left(g\right)\\ d,V_{ddAlCl_3}=V_{ddHCl}=0,3\left(l\right)\\ C_{MddHCl}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)

\(a.PTHH:Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\\ b.n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ PTHH:Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\\ n_{H_2}=0,2.2=0,4\left(mol\right)\\ V_{H_2}=0,4.22,4=8,96\left(l\right)\\ c.n_{HCl}=n_{Zn}=0,2mol\\ C_{MHCl}=\dfrac{0,4}{0,1}=4\left(M\right)\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH :

\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

0,2     0,4          0,2          0,2 

\(b,V_{H_2}=0,2.22,4=4,48\left(l\right)\)

\(c,C_M=\dfrac{n}{V}=\dfrac{0,4}{0,1}=4M\)

a. \(PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

\(n_{Mg}=\dfrac{m_{Mg}}{M_{Mg}}=\dfrac{12}{24}=0,5\left(mol\right)\)

- Mol theo PTHH : \(1:2:1:1\)

- Mol theo phản ứng : \(0,5\rightarrow1\rightarrow0,5\rightarrow0,5\)

\(\Rightarrow n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow V_{H_2}=n_{H_2}.22,4=0,5.22,4=11,2\left(l\right)\)

b. Từ a. \(\Rightarrow n_{HCl}=1\left(mol\right)\)

\(\Rightarrow m_{HCl}=n_{HCl}.M_{HCl}=1.\left(1+35,5\right)=36,5\left(g\right)\)

c. \(n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(PTHH:2H_2+O_2\underrightarrow{t^o}2H_2O\)

- Mol theo PTHH : \(2:1:2\)

- Mol theo phản ứng : \(0,6\leftarrow0,3\rightarrow0,6\)

\(\Rightarrow n_{H_2O}=0,6\left(mol\right)\)

\(\Rightarrow m_{H_2O}=n_{H_2O}.M_{H_2O}=0,6.\left(2+16\right)=10,8\left(g\right)\)

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

____0,1______0,2_____0,1____0,1 (mol)

a, \(C_{M_{HCl}}=\dfrac{0,2}{0,15}=\dfrac{4}{3}\left(M\right)\)

\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, \(FeCl_2+2NaOH\rightarrow2NaCl+Fe\left(OH\right)_2\)

Theo PT: \(n_{NaOH}=2n_{FeCl_2}=0,2\left(mol\right)\)

\(\Rightarrow V_{NaOH}=\dfrac{0,2}{2}=0,1\left(l\right)\)

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\) 
          0,1           0,2                     0,1 
\(V_{H_2}=0,1.22,4=2,24l\\ m_{HCl}=\dfrac{0,2.36,5}{10}.100=73g\)

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