a) CuO + 2HCl → CuCl2 + H2O (1)
Fe2O3 + 6HCl → 2FeCl3 + 3H2O (2)
\(n_{HCl}=0,2\times2,5=0,5\left(mol\right)\)
b) Gọi x,y lần lượt là số mol của CuO và Fe2O3
Ta có: \(80x+160y=16\) (*)
Theo Pt1: \(n_{HCl}=2n_{CuO}=2x\left(mol\right)\)
Theo pt2: \(n_{HCl}=6n_{Fe_2O_3}=6y\left(mol\right)\)
Ta có: \(2x+6y=0,5\) (**)
Từ (*)(**) ta có: \(\left\{{}\begin{matrix}80x+160y=16\\2x+6y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)
Vậy \(n_{CuO}=0,1\left(mol\right)\Rightarrow m_{CuO}=0,1\times80=8\left(g\right)\)
\(n_{Fe_2O_3}=0,05\left(mol\right)\Rightarrow m_{Fe_2O_3}=0,05\times160=8\left(g\right)\)
\(\%m_{CuO}=\%m_{Fe_2O_3}=\frac{8}{16}\times100\%=50\%\)
b) CuO + H2SO4 → CuSO4 + H2O (3)
Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O (4)
Theo Pt3: \(n_{H_2SO_4}=n_{CuO}=0,1\left(mol\right)\)
Theo pt4: \(n_{H_2SO_4}=3n_{Fe_2O_3}=3\times0,05=0,15\left(mol\right)\)
\(\Rightarrow\Sigma n_{H_2SO_4}=0,1+0,15=0,25\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,25\times98=24,5\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\frac{24,5}{20\%}=122,5\left(g\right)\)
Đặt \(n_{CuO}=x;n_{Fe_2O_3}=y\)
\(PTHH:CuO+2HCl\rightarrow CuCl_2+H_2O\)
(mol) 1 2 1 1
(mol) x 2x x x
\(PTHH:Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
(mol) 1 6 2 3
(mol) y 6y 2y 3y
\(hpt:\left\{{}\begin{matrix}80x+160y=16\\2x+6y=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n_{CuO}=0,1\left(mol\right)\\n_{Fe_2O_3}=0,05\left(mol\right)\end{matrix}\right.\)
\(m_{CuO}=80,0,1=8\left(g\right)\\ m_{Fe_2O_3}=16-8=8\left(g\right)\)
\(\%m_{CuO}=\frac{8}{16}.100=50\%\\ \%m_{Fe_2O_3}=100-50=50\%\)
\(PTHH:CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
(mol) 1 1
(mol) 0,1 0,1
\(PTHH:Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
(mol) 1 3
(mol) 0,05 0,15
\(m_{H_2SO_4}=98\left(0,1+0,15\right)=24,5\left(g\right)\)
\(m_{ddH_2SO_4}=\frac{24,5.100\%}{20\%}=122,5\left(g\right)\)
Đặt :
nCuO = x mol
nFe2O3 = y mol
<=> 80x + 160y = 16 (1)
nHCl = 0.5 mol
CuO + 2HCl --> CuCl2 + H2O
x_______2x
Fe2O3 + 6HCl --> 2FeCl3 + 3H2O
y________6y
<=> 2x + 6y = 0.5 (2)
Từ (1) và (2) :
x = 0.1
y = 0.05
mCuO = 8 g
mFe2O3 = 8 g
%CuO = %Fe2O3 = 50%
CuO + H2SO4 --> CuSO4 + H2O
0.1_____0.1
Fe2O3 + 3H2SO4 --> Fe2(SO4)3 + 3H2O
0.05______0.15
nH2SO4 = 0.25 mol
mH2SO4 = 24.5 g
mdd H2SO4 = 122.5 g
