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Ta có Fe có 3 hóa trị II và III
*TH1: Nếu Fe hóa trị II => CTDC:\(Fe_aX_2\)
\(Fe\) chiềm\(34,46\%\Rightarrow\frac{m_{Fe}}{m_{Fe_aX_2}}.100\%=34,46\%\)
\(\Leftrightarrow\frac{56a}{56a+2X}=0,3446\)
\(\Leftrightarrow56a=19,2976a+0,6892X\)
\(\Leftrightarrow X=54a\)
| a | 1 | 2 | 3 |
| X | 54(loại) | 108(loại) | 162(loại) |
*TH2: Fe hóa trị \(III\)
\(\Rightarrow CTDC:Fe_aX_3\)
\(Fe\) chiếm \(34,46\%\Rightarrow\frac{56a}{56a+3X}=0,3446\)
\(\Leftrightarrow56a=19,2976a+1,0338X\)
\(\Leftrightarrow X=35,5a\)
| a | 1 | 2 | 3 |
| X | 35,5(Cl) | 71(loại) | 106,5(loại) |
Vậy CTHH A là : \(FeCl_3\)
a) \(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=3n_{Fe_3O_4}=0,06\left(mol\right)\\n_O=4n_{Fe_3O_4}=0,08\left(mol\right)\end{matrix}\right.\)
b) \(n_{N_2O}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_N=2n_{N_2O}=0,3\left(mol\right)\\n_O=n_{N_2O}=0,15\left(mol\right)\end{matrix}\right.\)
c) Ta có: \(n_{H_2SO_4}=\dfrac{4,9}{98}=0,05\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_H=2n_{H_2SO_4}=0,1\left(mol\right)\\n_S=n_{H_2SO_4}=0,05\left(mol\right)\\n_O=4n_{H_2SO_4}=0,2\left(mol\right)\end{matrix}\right.\)
biết \(M_O=16\left(đvC\right)\)
\(\Rightarrow\)\(M_X=2,5.16=40\left(đvC\right)\)
\(\Rightarrow X\) là \(Ca\)
\(m_{Ca}=0,16605\times10^{-23}\times40=6,642\times10^{-23}\left(g\right)\)
\(m_{Mg}=0,16605\times10^{-23}\times24=3,9852\times10^{-23}\left(g\right)\)
\(m_{Al}=0,16605\times10^{-23}\times27=4,48335\times10^{-23}\left(g\right)\)
\(m_{Fe}=0,16605\times10^{-23}\times56=9,2988\times10^{-23}\left(g\right)\)
\(m_{Na}=0,16605\times10^{-23}\times23=3,81915\times10^{-23}\left(g\right)\)
\(m_O=0,16605\times10^{-23}\times16=2,6568\times10^{-23}\left(g\right)\)
\(m_S=0,16605\times10^{-23}\times32=5,3136\times10^{-23}\left(g\right)\)
\(m_N=0,16605\times10^{-23}\times14=2,3247\times10^{-23}\left(g\right)\)
\(m_{Cl}=0,16605\times10^{-23}\times35,5=5,894775\times10^{-23}\left(g\right)\)