\(n_{hh}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

\(m_{Br_2}=200\cdot\dfrac{20}{100}=40\left(g\right)\)

\(n_{Br_2}=\dfrac{40}{160}=0.25\left(mol\right)\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(0.25........0.25..........0.25\)

\(\)\(n_{C_2H_4}=n_{hh}=0.25\left(mol\right)\)

=> Sai đề

Bài 1 :

PTHH : Zn + H₂SO₄ ------> ZnSO₄ + H₂

\(n_{Zn}=\frac{m}{M}=\frac{26}{65}=0,4\left(mol\right)\)

Theo PTHH : n_H2 = n_Zn = 0,4 mol

=> Khối lượng H₂ được tạo ra ở đktc là :

\(V=n\times22,4\)

\(\Rightarrow V_{H_2}=0,4\times22,4\)

\(\Rightarrow V_{H_2}=8,96\left(l\right)\)

Theo PTHH : n_ZnSO4 = n_Zn = 0,4 mol

=> Khối lượng muối được tạo thành là :

\(m=n\times M\)

\(\Rightarrow m_{ZnSO_4}=0,4\times161\)

\(\Rightarrow m_{ZnSO_4}=64,4\left(g\right)\)

b) Theo PT : \(n_{H_2SO_4}=n_{Zn}=0,4\left(mol\right)\)

=> Khối lượng \(H_2SO_4\)cần dùng cho phản ứng là :

\(m=n\times M\)

\(\Rightarrow m_{H_2SO_4}=0,4\times98\)

\(\Rightarrow m_{H_2SO_4}=39,2\left(g\right)\)

c) Nồng độ phần trăm thu được sau phản ứng là :

\(C\%=\frac{m_{ct}}{m_{dd}}\times100\%\)

\(\Rightarrow C\%=\frac{39,2}{64,4}\times100\%\approx60,9\%\)

Vậy :.........................

SO2+Ba(OH)2->BaSO3+H2O

0,2-----0,2---------0,2--------0,2

n SO2=4,48\22,4=0,2 mol

=>Cm=0,2\0,25=0,8M

=>m BaSO3=0,2.217=43,4g

n Ba(OH)2 = 2.0,2=0,4 mol

  \(Ba\left(OH\right)_2+SO_2->BaSO_3+H_2O\)

0,4 ................0,4...........0,4

m BaSO3 = 0,4. ( 138+32+16.3)=87,2 g 

v SO2 = 0,4.22,4=8,96 lít

a)

PTHH : \(SO_2+Ca\left(OH\right)_2\rightarrow CáO_4+H_2O\)

b)

Ta có :

\(n_{SO_2}=\frac{0,224}{22,4}=0,01\left(mol\right)\)

\(n_{Ca\left(OH\right)_2}=0,01\times1,4=0,014\)

Theo ptpư : \(n_{SO_2}=n_{Ca\left(OH\right)_2}=n_{CaSO_3}=n_{H_2O}\)

Vậy n_{Ca(OH)2 ( dư )} = \(n_{Ca\left(OH\right)_2\left(bđ\right)}-n_{Ca\left(OH\right)_2\left(pư\right)}\)

\(=0,014-0,001=0,004\left(mol\right)\)

\(m_{NaOH}=\dfrac{200\cdot8}{100}=16\left(g\right)\Rightarrow n_{NaOH}=\dfrac{16}{40}=0,4mol\)

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

 0,4           0,4         0,4        0,4

a)\(m_{HCl}=0,4\cdot36,5=14,6\left(g\right)\) 

 \(\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3}\cdot100=200\left(g\right)\)

b)\(m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\)

   \(m_{H_2O}=0,4\cdot18=7,2\left(g\right)\)

   \(m_{ddsau}=200+200-7,2=392,8\left(g\right)\)

   \(\Rightarrow C\%=\dfrac{23,4}{392,8}\cdot100=5,96\%\)

c) \(n_{SO_2}=\dfrac{6,72}{22,4}=0,3mol\)

     \(2NaOH+SO_2\rightarrow Na_2SO_4+H_2O\)

         0,4          0,3         0,3              0,3

     \(m_{Na_2SO_4}=0,3\cdot142=42,6\left(g\right)\)

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

Theo PT:  \(n_{Ca\left(OH\right)_2}=n_{CaCO_3}=n_{CO_2}=0,25\left(mol\right)\)

a, \(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)

b, \(m_{CaCO_3}=0,25.100=25\left(g\right)\)

c, \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

Theo PT: \(n_{HCl}=2n_{Ca\left(OH\right)_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{20\%}=91,25\left(g\right)\)

\(a.n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{17,8\%.200}{36,5}=\dfrac{356}{365}\left(mol\right)\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,1}{1}< \dfrac{\dfrac{356}{365}}{2}\\ \Rightarrow Znhết,HCldư\\ n_{HCl\left(dùng\right)}=0,1.2=0,2\left(mol\right)\\ m_{HCl\left(dùng\right)}=0,2.36,5=7,3\left(g\right)\\ b.n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c.n_{HCl\left(Dư\right)}=\dfrac{356}{365}-0,2=\dfrac{283}{365}\left(mol\right)\\ C\%_{ddZnCl_2}=\dfrac{0,1.136}{6,5+200}.100\approx6,586\%\)

\(C\%_{ddHCl\left(dư\right)}=\dfrac{\dfrac{283}{365}.36,5}{6,5+200}.100\approx13,705\%\)