Fe tác dụng với HNO3 sẽ sinh ra Fe³⁺ em nha do HNO₃ có tính oxi hóa mạnh nên sẽ đẩy Fe lên mức cao nhất

Sơ đồ:

\(24,5g\left(A\right)\left\{{}\begin{matrix}Fe\\Cu\\Ag\end{matrix}\right.\xrightarrow[1,6M]{+HNO_3}\left(B\right)\left\{{}\begin{matrix}Fe^{3+}\\Cu^{2+}\\Ag^+\\NO_3^-\end{matrix}\right.+NO:6,372l\)

\(n_{NO}=\dfrac{6,272}{22,4}=0,28mol\\ BTe:3n_{Fe}+2n_{Cu}+n_{Ag}=3n_{NO}=3.0,28=0,84mol\\ m=24,5+0,84.62=76,58g\)

\(BTNT\left(N\right):n_{HNO_3}=n_{NO_3^-}+n_{NO}\\ \Leftrightarrow n_{HNO_3}=0,84+0,28\\ \Leftrightarrow n_{HNO_3}=1,12mol\\ \\ V=V_{HNO_3}=\dfrac{1,12}{1,6}=0,7l=700ml\)

Gọi CTHH oxit kim loại là \(RO\)

Giả sử có 1mol oxit pứ

\(RO+H_2SO_4\rightarrow RSO_4+H_2O\)

\(1-\rightarrow1---\rightarrow1\)

\(m_{ddH_2SO_4}=\dfrac{1.98}{4,9}\cdot100=2000\left(g\right)\\ m_{ddRSO_4}=1\left(R+16\right)+2000=2016+R\left(g\right)\\ C_{\%RSO_4}=\dfrac{1\left(R+96\right)}{2016+R}\cdot100=5,88\%\\ \Rightarrow R\approx24\left(g/mol\right)\)

Vậy R là Mg

Giỏi Hóa quas ò 🏆

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Al_2\left(SO_4\right)_3+3BaCl_2\rightarrow3BaSO_4+2AlCl_3\)

Ta có: n_N2 = 0,22 (mol)

⇒ n_NO3^- = 10n_N2 = 2,2 (mol)

⇒ m _muối = m_X + m_NO3^- = 31 + 2,2.62 = 167,4 (g)

n_HNO3 = 12n_N2 = 2,64 (mol)

\(\Rightarrow V_{HNO_3}=\dfrac{2,64}{2}=1,32\left(l\right)=1320\left(ml\right)\)

\(H_2SO_4+BaCl_2\rightarrow BaSO_4+2HCl\\ BaCl_2+Na_2SO_4\rightarrow BaSO_4+2NaCl\)

\(a)n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ 0,2\leftarrow-0,3\leftarrow-0,1\leftarrow---0,3\)

\(a=m_{Al}=0,2.27=5,4g\\ b)m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\\ c)C_{\%H_2SO_4}=\dfrac{0,3.98}{100}\cdot100=29,4\%\)

a)nH2​​=22,46,72​=0,3mol2Al+3H2​SO4​→Al2​(SO4​)3​+3H2​0,2←−0,3←−0,1←−−−0,3

$a=m^{Al}=0,2.27=5,4gb)m^{A{l}^2{\left(S{O}^4\right)}^3}=0,1.342=34,2gc)C^{\%H^{2}S{O}^4}=\genfrac{}{}{0ex}{}{0,3.98}{100}\cdot 100=29,4\%$

\(Pt: 2Al+3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3H_2\)

\(a.n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo pt: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Al}=a=0,2.27=5,4\left(g\right)\)

\(b.n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\)

\(c.\)Theo pt: \(n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,3.98=29,4g\)

\(C_{\%}H_2SO_4=\dfrac{29,4}{100}.100\%=29,4\%\)

\(Pt: Fe + 2HCl \rightarrow FeCl_2 + H_2\)

\(a.n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Theo pt: \(nH_2 = nFe = 0,2 mol\)

\(\Rightarrow V_{H_2}=0,2.22,4=4,48l\)

\(b.n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=0,2.127=25.4g\)

\(c.n_{HCl}=2nFe=0,4mol\)

\(C_MHCl=\dfrac{0,4}{0,1}=4M\)

\(a)n_{Fe}=\dfrac{11,2}{56}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ 0,2\rightarrow0,4-\rightarrow0,2-\rightarrow0,2\)

\(V_{H_2}=0,2.22,4=4,48l\\ b)m_{FeCl_2}=0,2.127=25,4g\\ c)C_{M_{HCl}}=\dfrac{0,4}{0,1}=4M\)