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

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

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

\(n_{HCl}=0,4.2=0,8\left(mol\right)\\ Đặt:n_{CuO}=a\left(mol\right);n_{Fe_2O_3}=b\left(mol\right)\left(a,b>0\right)\\CuO+2HCl\rightarrow CuCl_2+H_2O\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O \\ \Rightarrow\left\{{}\begin{matrix}80a+160b=24\\2a+6b=0,8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ C1:m_{muối}=135a+2.162,5b=135.0,1+325.0,1=46\left(g\right)\\ C2:n_{H_2O}=\dfrac{n_{HCl}}{2}=0,4\left(mol\right)=n_{O\left(trongH_2O\right)}=n_{O\left(trong.oxit\right)}\\ \Rightarrow m_{muối}=m_{CuO,Fe_2O_3}-m_{O\left(trong.oxit\right)}+m_{Cl^-}=24-0,4.16+0,8.35,5=46\left(g\right)\\ b,m_{CuO}=80a=8\left(g\right);m_{Fe_2O_3}=160b=16\left(g\right)\)

Làm vài bài rồi ra sân bay check in đây

CuO+2HCl->CuCl₂+H₂O

x--------2x

Fe₂O₃+6HCl->2FeCl₃+3H₂O

y--------------6y

Ta  có :

\(\left\{{}\begin{matrix}56x+160y=24\\2x+6y=0,8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

=>m _CuO=0,1.56=5,6g

=>m _Fe2O3=0,1.160=16g

-> m _muối =0,1.135+0,2.162,5=46g

C2:n _H2O=0,4 mol

=>m _muối =24-0,4.16+0,8.35,5=46g

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(CaO+2HCl\rightarrow CaCl_2+H_2O\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(n_{HCl}=0,6.2=1,2\left(mol\right)\)

\(m_{HCl}=1,2.36,5=43,8\left(g\right)\)

\(m_{H_2O}=0,6.18=10,8\left(g\right)\)

\(m_{muoi.khan}=37,6+43,8-10,8=70,6\left(g\right)\) ( định luật bảo toàn khối lượng )

Ta có:

Tương tự bài 1, ta có:

m 5 o x i t + m H C l = m m u o i + m H 2 O   s a n   p h a m

m m u o i = m 5 o x i t + m H C l - m H 2 O   s a n   p h a m

⇔ m m u o i  = 7,2g

⇒ Chọn C.

PT :\(\left(1\right)NaOH+HCl\rightarrow NaCl+H_2O\)

x mol \(\rightarrow\) x mol

\(\left(2\right)KOH+HCl\rightarrow KCl+H_2O\)

y mol \(\rightarrow\) y mol

Mà ta có :40x + 57y = 15,2

58,5x + 75,5y = 20,75

Giải hệ phương trình có :x=\(\dfrac{19}{170}mol;y=\dfrac{16}{85}mol\)

\(m_{NaOH}=\dfrac{19}{170}.40=4,47g\)

\(\%m_{NaOH}=\dfrac{4,47}{15,2}.100=29,4\%\).

\(\Rightarrow\%m_{KOH}=70,6\%\)

b,

Ta có: \(m_{HCl}=36,5.\dfrac{19}{170}+36,5.\dfrac{16}{85}=43,48g\)

\(m_{dd_{NaOH}}=\dfrac{43,48.100}{7,3}=595,6\%\).

Vậy \(C_{\%_{ddthu}}=\dfrac{20,75}{595,6}.100=3,48\%\) .

a, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:     0,1                    0,1      0,1

PTHH: CuO + 2HCl → CuCl₂ + H₂O

Mol:       0,01                    0,01

\(\%m_{Zn}=\dfrac{0,1.65.100\%}{7,3}=89,04\%\)

\(\%m_{CuO}=100-89,04=10,96\%\)

b, \(n_{CuO}=\dfrac{7,3-0,1.65}{80}=\dfrac{0,8}{80}=0,01\left(mol\right)\)

\(m_{muối}=0,1.136+0,01.135=14,95\left(g\right)\)

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

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

 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

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

PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

Ta có: \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,1\left(mol\right)\\n_{CuCl_2}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,1\cdot36,5}{7,3\%}=50\left(g\right)\\C\%_{CuCl_2}=\dfrac{0,05\cdot135}{4+50}\cdot100\%=12,5\%\end{matrix}\right.\)

 Tương tự bài 1, ta có:

m 4 o x i t = m m u o i - m H 2 S O 4 + m H 2 O   s a n   p h a m

m 4 o x i t  = 3,61g

⇒ Chọn C.