\(Fe_2O_3 + 6HCl \rightarrow 2FeCl_3 + 3H_2O\)

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

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

\(n_{NaOH} = 0,2 . 1 = 0,2 mol\)

\(n_{HCl dư} = n_{NaOH}= 0,2 mol\)

\(\Rightarrow n_{HCl pư}= n_{HCl ban đầu} - n_{HCl dư}= 1,4- 0,2 = 1,2 mol\)

Gọi n_{\(Fe_2O_3\)} và n_\(CuO\) là x, y

\(\begin{cases} 160x + 80y=40\\ 6x + 2y= 1,2 \end{cases} \)

\(\begin{cases} x=0,1\\ y=0,3 \end{cases} \)

\(\Rightarrow m_{Fe_2O_3}= 0,1 . 160= 16g\)

\(m_{CuO} = 0,3 . 80=24g\)

\(n_{HCl}=0,5.1=0,5\left(mol\right)\)

a)

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

x------->2x

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

y--------->6y

Có hệ: \(\left\{{}\begin{matrix}2x+6y=0,5\\80x+160y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

\(m_{CuO}=0,1.80=8\left(g\right)\\ m_{Fe_2O_3}=0,05.160=8\left(g\right)\)

b

\(\%m_{CuO}=\dfrac{0,1.80.100\%}{16}=50\%\\ \%m_{Fe_2O_3}=\dfrac{0,05.160.100\%}{16}=50\%\)

Gọi \(\left\{{}\begin{matrix}n_{CuO}=x\\n_{MgO}=y\end{matrix}\right.\)

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

   x                           x                   ( mol )

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

 y                            y                  ( mol )

Ta có:

\(\left\{{}\begin{matrix}80x+40y=16\\135x+95y=32,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(\Rightarrow m_{CuO}=0,1.80=8g\)

\(\Rightarrow m_{MgO}=0,2.40=8g\)

\(\%m_{CuO}=\dfrac{8}{16}.100=50\%\)

\(\%m_{MgO}=\dfrac{8}{16}.100=50\%\)

\(m_{CuCl_2}=0,1.135=13,5g\)

\(m_{MgCl_2}=0,2.95=19g\)

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

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

b, Gọi: \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\) ⇒ 80x + 160y = 32 (1)

Theo PT: \(\left\{{}\begin{matrix}n_{CuCl_2}=n_{Cu}=x\left(mol\right)\\n_{FeCl_3}=2n_{Fe_2O_3}=2y\left(mol\right)\end{matrix}\right.\) ⇒ 135x + 325y = 59,5 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,2.80=16\left(g\right)\\m_{Fe_2O_3}=0,1.160=16\left(g\right)\end{matrix}\right.\)

c, Theo PT: \(n_{HCl}=2n_{CuO}+6n_{Fe_2O_3}=1\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{1}{0,5}=2\left(l\right)\)

dd chi đó bn

a,\(n_{HCl}=0,5.1=0,5\left(mol\right)\)

PTHH: CuO + 2HCl → CuCl₂ + H₂O

Mol:       x          2x

PTHH: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

Mol:        y            6y

Ta có: \(\left\{{}\begin{matrix}80x+160y=16\\2x+6y=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

PTHH: CuO + 2HCl → CuCl₂ + H₂O

Mol:      0,1       0,2

PTHH: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

Mol:      0,05         0,3

\(\Rightarrow m_{CuO}=0,1.80=8\left(g\right);m_{Fe_2O_3}=16-8=8\left(g\right)\)

b,\(\%m_{CuO}=\dfrac{8.100\%}{16}=50\%;\%m_{Fe_2O_3}=100-50=50\%\)

Bài 1:

Ta có: \(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)

BTNT H, có: \(n_{HCl}=2n_{H_2O}\Rightarrow n_{H_2O}=0,05\left(mol\right)\)

Theo ĐL BTKL, có: m _oxit + m_HCl = m_muối + m_H2O

⇒ m_muối = 2,8 + 0,1.36,5 - 0,05.18 = 5,55 (g)

Bài 2:

\(m_{KOH}=200.5,6\%=11,2\left(g\right)\Rightarrow n_{KOH}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PT: \(2KOH+CuCl_2\rightarrow2KCl+Cu\left(OH\right)_2\)

Theo PT: \(n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{KOH}=0,1\left(mol\right)\)

\(\Rightarrow m_{Cu\left(OH\right)_2}=0,1.98=9,8\left(g\right)\)

Bài 3:

Ta có: \(C\%_{NaCl}=\dfrac{m_{NaCl}}{m_{NaCl}+m_{H_2O}}.100\%=32\%\)

\(\Rightarrow\dfrac{m_{NaCl}}{m_{NaCl}+200}=0,32\)

\(\Rightarrow m_{NaCl}\approx94,12\left(g\right)\)

Bài 4:

Ta có: \(n_{CaCO_3}=\dfrac{15}{100}=0,15\left(mol\right)\)

PT: \(CaCO_3\underrightarrow{t^o}CaO+CO_2\)

Theo PT: \(n_{CaO\left(LT\right)}=n_{CaCO_3}=0,15\left(mol\right)\)

\(\Rightarrow m_{CaO\left(LT\right)}=0,15.56=8,4\left(g\right)\)

\(\Rightarrow H=\dfrac{6,72}{8,4}.100\%=80\%\)