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a, \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Theo PT: \(n_{H_2}=n_{Fe}=0,4\left(mol\right)\Rightarrow V_{H_2}=0,4.22,4=8,96\left(l\right)\)
b, \(n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,25}{1}\), ta được Fe dư.
Theo PT: \(n_{Fe\left(pư\right)}=n_{H_2SO_4}=0,25\left(mol\right)\Rightarrow n_{Fe\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)
\(\Rightarrow m_{Fe\left(dư\right)}=0,15.56=8,4\left(g\right)\)
1
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\
pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
0,1 0,1 0,1 0,1
\(V_{H_2}=0,1.22,4=2,24\left(l\right)\\
m_{FeSO_4}=127.0,1=12,7\left(g\right)\)
\(m_{\text{dd}}=5,6+500-\left(0,1.2\right)=505,4\left(g\right)\\
C\%_{FeSO_4}=\dfrac{12,7}{505,4}.100\%=2,513\%\)
2
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\
pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
0,2 0,2 0,2 0,2
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\
m_{MgSO_4}=120.0,2=24\left(g\right)\\
V_{\text{dd}H_2SO_4}=\dfrac{0,2}{1}=0,2M\)
Câu 4:
a, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
\(4K+O_2\underrightarrow{t^O}2K_2O\)
b, \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
c, \(2K+2H_2O\rightarrow2KOH+H_2\)
\(CaO+H_2O\rightarrow Ba\left(OH\right)_2\)
\(SO_3+H_2O\rightarrow H_2SO_4\)
Do nồi nhôm bị oxi hóa bởi oxi trong không khí tạo thành lớp màng oxit bên ngoài bề mặt nhôm :
\(4Al + 3O_2 \to 2Al_2O_3\)
MFe2O3=56×2+16×3=160(g/mol)MFe2O3=56×2+16×3=160(g/mol)
%mFe=56×2160.100%=70%%mFe=56×2160.100%=70%
MFe3O4=56×3+16×4=232(g/mol)MFe3O4=56×3+16×4=232(g/mol)
%mFe=56×3232.100%≈72,4%
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