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

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

 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

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

ra Na2SO3 mà bạn

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

a) \(n_{CO_2}=\dfrac{V}{22,4}=\dfrac{3,36}{22,3}=0,15\left(mol\right)\)

\(\Rightarrow n_{NaOH}=2n_{CO_2}=0,3\left(mol\right)\)

\(C_{M_{ddNaOH}}=\dfrac{n}{V}=\dfrac{0,3}{0,3}=1M\)

b) \(n_{Na_2CO_3}=n_{CO_2}=0,15\left(mol\right)\)

\(m_{Na_2CO_3}=n.M=0,15.106=15,9\left(g\right)\)

Ta có: \(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

a, \(n_{Fe}=n_{H_2}=0,45\left(mol\right)\Rightarrow m_{Fe}=0,45.56=25,2\left(g\right)\)

b, \(n_{HCl}=2n_{H_2}=0,9\left(mol\right)\) \(\Rightarrow C_{M_{HCl}}=\dfrac{0,9}{0,15}=6\left(M\right)\)

c, \(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)

\(4Fe\left(OH\right)_2+O_2\underrightarrow{t^o}2Fe_2O_3+4H_2O\)

Theo PT: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_2}=\dfrac{1}{2}n_{FeCl_2}=\dfrac{1}{2}n_{H_2}=0,225\left(mol\right)\)

\(\Rightarrow m_{Fe_2O_3}=0,225.160=36\left(g\right)\)

cảm ơn nha 

\(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1mol\)

\(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+SO_2+H_2O\)

a) \(n_{SO_2}=n_{Na_2SO_3}=0,1mol\) \(\Rightarrow V=2,24l\)

b) \(n_{H_2SO_4}=n_{Na_2SO_3}=0,1mol\) \(\Rightarrow C_M=\dfrac{0,1}{0,2}=0,5M\)

c) \(m_{Na_2SO_4}=0,1\cdot142=14,2g\)

Chị làm cụ thể câu c cho em được khong ạ, sao nNa2SO4 lại =0,1 mol thế chị

\(n_{khi}=\frac{2,24}{22.4}=0,1mol\)

ta có CaCl₂không tác dụng vớ HCl

PTHH: CaSO₃+HCl---> CaCl₂+H₂O+SO₂

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

=> m(CaSO₃)=0,1( 40+32+16.3)=12g

đề sai k bạn

1.\(n_{CO_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(n_{KOH}=CM.Vdd=2.0,46=0,92\left(mol\right)\)

PTHH:\(CO_2+2KOH\rightarrow K_2CO_3+H_2O\)

TPU:   0,2          0,92

PU:     0,2           0,4            0,2

SPU:    0             0,52            0,2

\(m_{K_2CO_3}=n.M=0,2.138=27,6\left(g\right)\)

2.\(n_{SO_2}=\dfrac{V}{22,4}=\dfrac{8,69}{22,4}=0,4\left(mol\right)\)

\(n_{NaOH}=CM.Vdd=2,5.0,2=0,5\left(mol\right)\)

PTHH:\(SO_2+2NaOH\rightarrow Na_2SO_3+H_2O\)

TPU:    0,4         0,5

PU:       0,25       0,5            0,25

SPU:    0,15        0               0,25

\(m_{Na_2SO_3}=n.M=0,25.126=31,5\left(g\right)\)

a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

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

Theo PT: \(n_{FeSO_4}=n_{H_2SO_4}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{FeSO_4}=0,2.152=30,4\left(g\right)\)

c, \(C_{M_{H_2SO_4}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)

\(a)Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ b)n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\\ n_{FeSO_4}=n_{H_2SO_4}=n_{H_2}=0,2mol\\ m_{FeSO_4}=0,2.152=30,4g\\ c)C_{M_{H_2SO_4}}=\dfrac{0,2}{0,5}=0,4M\)