\(m_{1đ.v.C}=\dfrac{1,9926.10^{-23}}{12}=0,16605.10^{-23}\left(g\right)\\ m_S=32.0,16605.10^{-23}=5,3136.10^{-23}\left(g\right)\)

- Tổng số hạt cơ bản của nguyên tử A là 40: P+N=E= 40 (1)

- Nguyên tử A trung hoà về điện: P=E=Z (2)

- Mặt khác: N-P=1 (3)

(1), (2), (3) lập hệ:

\(\left\{{}\begin{matrix}P+E+N=40\\P=E=Z\\N-P=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}P=E=Z=13\\N=14\end{matrix}\right.\)

Với Z=13 => A là Nhôm (Z_Al=13)

Cấu hình A: 1s²2s²2p⁶3s²3p¹

Sơ đồ nguyên tử A: 

\(n_{Cu}=\dfrac{9,6}{64}=0,15\left(mol\right)\\3 Cu+8HNO_3\rightarrow3Cu\left(NO_3\right)_2+2NO+4H_2O\\ n_{HNO_3}=\dfrac{8}{3}.0,15=0,4\left(mol\right)\\ n_{Cu\left(NO_3\right)_2}=n_{Cu}=0,15\left(mol\right)\\ n_{NO}=\dfrac{2}{3}.0,15=0,1\left(mol\right)\\ V=V_{NO\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ m_{Cu\left(NO_3\right)_2}=188.0,15=28,2\left(g\right)\\ m_{HNO_3}=63.0,4=25,2\left(g\right)\)

\(Đặt.CTTQ:Fe_aO_b\left(a,b:nguyên,dương\right)\\ Ta.có:a=\dfrac{70\%.160}{56}=2\\ b=\dfrac{160-2.56}{16}=3\\ Với:a=2;b=3\Rightarrow CTHH:Fe_2O_3\left(Sắt\left(III\right)oxit\right)\)

Cảm ơn rất nhiều ah

\(Đặt:a=\%V_{H_2}\left(a>0\right)\\\overline{M}_{hhA}=8,5.2=17\left(\dfrac{g}{mol}\right)\\ \Leftrightarrow 2a+32.\left(100\%-a\right)=17\\ \Leftrightarrow a=50\%\\ \Rightarrow\%V_{H_2}=\%V_{O_2}=50\%\\ \%m_{H_2}=\dfrac{2}{2+32}.100\approx5,882\%\Rightarrow\%m_{O_2}\approx100\%-5,882\%\approx94,118\%\)

2KMnO₄-to>K2MnO₄+MnO₂+O₂

1------------------------------------0,5

m _O2=16 g

->n_O2=\(\dfrac{16}{32}\)=0,5 mol

=>%m KMnO4=\(\dfrac{1.158}{316}.100=50\%\)

\(2KMnO_4\underrightarrow{^{^{t^0}}}K_2MnO_4+MnO_2+O_2\)

Bảo toàn khối lượng : 

\(m_{O_2}=m_{KMnO_4}-m_{Cr}=316-300=16\left(g\right)\)

\(n_{O_2}=\dfrac{16}{32}=0.5\left(mol\right)\)

\(\Rightarrow n_{KMnO_4}=0.5\cdot2=1\left(mol\right)\)

\(\%KMnO_{4\left(np\right)}=\dfrac{1\cdot158}{300}\cdot100\%=52.67\%\)

\(\overline{M}_{hhX}=\dfrac{2.44+3.16}{2+3}=27,2\left(\dfrac{g}{mol}\right)\\ \Rightarrow d_{\dfrac{X}{H_2}}=\dfrac{27,2}{2}=13,6\)

Cách giải này không ổn !

cj nhắn tin e có j hong ạ

\(n_{Fe}=a\left(mol\right),n_{Al}=b\left(mol\right)\)

\(m_{hh}=56a+27b=11\left(g\right)\left(1\right)\)

\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(a.........a..............a\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(b.........1.5b................1.5b\)

\(n_{H_2}=a+1.5b=0.4\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.1,b=0.2\)

\(\%Fe=\dfrac{0.1\cdot56}{11}\cdot100\%=50.91\%\)

\(\%Al=100\%-50.91\%=49.09\%\)

\(n_{H_2SO_4}=0.4\cdot2=0.8\left(mol\right)\)

\(C\%_{H_2SO_4}=\dfrac{0.8\cdot98}{200}\cdot100\%=39.2\%\)

dạ em cảm ơn

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