1) Fe2O3 + 3H2 \(\underrightarrow{to}\) 2Fe + 3H2O (1)
CuO + H2 \(\underrightarrow{to}\) Cu + H2O (2)
\(n_{H_2}=\frac{8,96}{22,4}=0,4\left(mol\right)\)
Gọi x,y lần lượt là số mol của Fe2O3 và CuO
Ta có: \(160x+80y=24\) (*)
Theo Pt1: \(n_{H_2}=3n_{Fe_2O_3}=3x\left(mol\right)\)
Theo Pt2: \(n_{H_2}=n_{CuO}=y\left(mol\right)\)
Ta có: \(3x+y=0,4\) (**)
Từ (*)(**) ta có: \(\left\{{}\begin{matrix}160x+80y=24\\3x+y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
Vậy \(n_{Fe_2O_3}=0,1\left(mol\right)\Rightarrow m_{Fe_2O_3}=0,1\times160=16\left(g\right)\)
\(n_{CuO}=0,1\left(mol\right)\Rightarrow m_{CuO}=0,1\times80=8\left(g\right)\)
\(\%m_{Fe_2O_3}=\frac{16}{24}\times100\%=66,67\%\)
\(\%m_{CuO}=100\%-66,67\%=33,33\%\)
2) Fe2O3 + 3CO \(\underrightarrow{to}\) 2Fe + 3CO2 (3)
CuO + CO \(\underrightarrow{to}\) Cu + CO2 (4)
a) Theo PT3: \(n_{CO}=3n_{Fe_2O_3}=3\times0,1=0,3\left(mol\right)\)
Theo pT4: \(n_{CO}=n_{CuO}=0,1\left(mol\right)\)
\(\Rightarrow\Sigma n_{CO}=0,3+0,1=0,4\left(mol\right)\)
\(\Rightarrow V_{CO}=0,4\times22,4=8,96\left(l\right)\)
b) Theo pT3: \(n_{Fe}=2n_{Fe_2O_3}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,2\times56=11,2\left(g\right)\)
Theo pT4: \(n_{Cu}=n_{CuO}=0,1\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,1\times64=6,4\left(g\right)\)
\(\%m_{Fe}=\frac{11,2}{11,2+6,4}\times100\%=63,64\%\)
\(\%m_{Cu}=100\%-63,64\%=36,36\%\)
1)
Đặt : nFe2O3= x mol
nCuO = y mol
nCO2 =
Fe2O3 + 3H2-to-> 2Fe + 3H2O (1)
x________3x
CuO + H2 -to-> Cu + H2O (2)
y_____y
mhh= 160x + 80y = 24 g (I)
nH2 = 3x + y = 0.4 mol (II)
Từ (I) và (II) :
=> x = y = 0.1
mFe2O3 = 16g
mCuO = 8g
%Fe2O3 = 66.67%
%CuO = 33.33%
2) Fe2O3 + 3CO -to-> 2Fe + 3CO2 (3)
CuO + CO -to-> Cu + CO2 (4)
Từ (1), (2) , (3), (4) :
nCO = nH2 = 0.4 mol
VCO = 0.4*22.4 = 8.96l
Từ (3) => nFe = 2nFe2O3 = 0.2 mol => mFe = 11.2g
Từ (4) => nCu = nCuO = 0.1 mol => mCu = 6.4 g
%Fe = 11.2/(11.2+6.4)*100% = 63.64%
%Cu= 36.36%
