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

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

 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

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

A là gì đấy nhỉ ?

a hả 

a là khoa 2k7 và là một streamer nimo về game miniworld 

PTHH: \(K_2O+2HCl\rightarrow2KCl+H_2O\)

            \(K_2SO_3+2HCl\rightarrow2KCl+SO_2\uparrow+H_2O\)

a) Ta có: \(\left\{{}\begin{matrix}n_{HCl}=\dfrac{200\cdot14,6\%}{36,5}=0,8\left(mol\right)\\n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{K_2SO_3}=0,3\left(mol\right)\\n_{K_2O}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{K_2O}=\dfrac{0,1\cdot94}{0,1\cdot94+0,3\cdot158}\cdot100\%\approx16,55\%\\\%m_{K_2SO_3}=83,45\%\end{matrix}\right.\)

b) Theo các PTHH: \(n_{KCl}=0,8\left(mol\right)\) \(\Rightarrow m_{KCl}=74,5\cdot0,8=59,6\left(g\right)\)

Mặt khác: \(\left\{{}\begin{matrix}m_{hh}=56,8\left(g\right)\\m_{SO_2}=0,3\cdot64=19,2\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{SO_2}=237,6\left(g\right)\)

\(\Rightarrow C\%_{KCl}=\dfrac{59,6}{237,6}\cdot100\%\approx25,1\%\)

\(a,PTHH:CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2\downarrow+2KCl\\ n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\\ \Rightarrow n_{KOH}=2n_{CuCl_2}=0,4\left(mol\right)\\ \Rightarrow C\%_{KOH}=\dfrac{0,4}{0,2}=2M\\ b,PTHH:Cu\left(OH\right)_2\rightarrow^{t^o}CuO+H_2O\\ \Rightarrow n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,2\left(mol\right)\\ \Rightarrow m_{CuO}=0,2\cdot80=16\left(g\right)\)

CH3COOH+NaHCO3->CH3COONa+H2O+CO2

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

n CO2=0,1 mol

=>C% axit=\(\dfrac{0,1.60}{200}.100\)=3%

m NaHCO3=0,1.84=8,4g

c) C% CH3COONa=\(\dfrac{0,1.82}{200+8,4}.100=3,934\%\)

Câu 3 : 

\(m_{ct}=\dfrac{10.80}{100}=8\left(g\right)\)

\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

a) Hiện tượng : Xuất hiện kết tủa trắng

Pt : \(2NaOH+MgSO_4\rightarrow Na_2SO_4+Mg\left(OH\right)_2|\)

           2                1                  1               1

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

\(n_{Mg\left(OH\right)2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{Mg\left(OH\right)2}=0,1.58=5,8\left(g\right)\)

b) \(n_{MgSO4}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

 \(m_{MgSO4}=0,1.120=12\left(g\right)\)

\(m_{ddMgSO}=\dfrac{12.100}{10}=120\left(g\right)\)

c) \(n_{Na2SO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

⇒ \(m_{Na2SO4}=0,1.142=14,2\left(g\right)\)

\(m_{ddspu}=80+120-5,8=194,2\left(g\right)\)

\(C_{Na2SO4}=\dfrac{14,2.100}{194,2}=7,31\)⁰/₀

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Câu 4 : 

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2|\)

         1         1               1           1

        0,2      0,2

\(n_{H2SO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(m_{H2SO4}=0,2.98=19,6\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{19,6.100}{20}=98\left(g\right)\)

\(V_{ddH2SO4}=\dfrac{98}{1,2}\simeq81,67\left(ml\right)\)

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