\(m_{H_2SO_4}=\dfrac{200\cdot9,8\%}{100\%}=19,6\left(g\right)\\ n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\\ PTHH:2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\\ \Rightarrow n_{KOH}=2n_{H_2SO_4}=0,4\left(mol\right)\\ \Rightarrow m_{KOH}=0,4\cdot\left(39+16+1\right)=22,4\left(g\right)\)

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

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

 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

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

ra Na2SO3 mà bạn

\(m_{Ba\left(OH\right)_2}=\dfrac{100\cdot17,1\%}{100\%}=17,1\left(g\right)\\ n_{Ba\left(OH\right)_2}=\dfrac{17,1}{171}=0,1\left(mol\right)\\ PTHH:Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+H_2O\\ n_{BaCl_2}=n_{Ba\left(OH\right)_2}=0,1\left(mol\right)\\ \Rightarrow m_{muối.sau.p/ứ}=m_{BaCl_2}=0,1\cdot208=20,8\left(g\right)\)

\(n_{Na_2CO_3}=0,1.1=0,1\left(mol\right)\)

a. \(Na_2CO_3+Ba\left(OH\right)_2\rightarrow BaCO_3+2NaOH\)

       0,1            0,1                   0,1           0,2

b. \(m_{kt}=m_{BaCO_3}=0,1.197=19,7\left(g\right)\)

c. \(C\%_{Ba\left(OH\right)_2}=\dfrac{0,1.171.100}{200}=8,55\%\)

d. \(BaCO_3+2HCl\rightarrow BaCl_2+H_2O+CO_2\)

      0,1           0,2

=> \(a=m_{dd.HCl}=\dfrac{0,2.36,5.100}{30}=\dfrac{73}{3}\left(g\right)\)

PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

Ta có: \(n_{Fe_2O_3}=\dfrac{4,8}{160}=0,03\left(mol\right)\) 

\(\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=0,09\left(mol\right)\\n_{Fe_2\left(SO_4\right)_3}=0,03\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,09\cdot98}{9,8\%}=90\left(g\right)\\m_{Fe_2\left(SO_4\right)_3}=0,03\cdot400=12\left(g\right)\end{matrix}\right.\)

a) 

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

PTHH: Na₂CO₃ + 2CH₃COOH --> 2CH₃COONa + CO₂ + H₂O

               0,15<---------0,3<------------------------0,15

=> \(C\%_{dd.CH_3COOH}=\dfrac{0,3.60}{200}.100\%=9\%\)

b)

\(m_{dd.Na_2CO_3}=\dfrac{0,15.106.100}{15}=106\left(g\right)\)

c)

PTHH: 2CH₃COOH + Ba(OH)₂ --> (CH₃COO)₂Ba + 2H₂O

                    0,3--------->0,15

=> \(V_{dd.Ba\left(OH\right)_2}=\dfrac{0,15}{0,5}=0,3\left(l\right)=300\left(ml\right)\)

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

Cho e hỏi công thức dùng để tính ra số mol KOH