a, PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

Gọi: \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\) ⇒ 80x + 160y = 11,2 (1)

Ta có: \(m_{HCl}=146.10\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{CuO}+6n_{Fe_2O_3}=2x+6y=0,4\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,02\left(mol\right)\\y=0,06\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,02.80}{11,2}.100\%\approx14,29\%\\\%m_{Fe_2O_3}\approx85,71\%\end{matrix}\right.\)

b, PT: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

Theo PT: \(n_{H_2SO_4}=n_{CuO}+3n_{Fe_2O_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,2.98=19,6\left(g\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{19,6}{4,9\%}=400\left(g\right)\)

Dung dịch A tác dụng với $Ba(OH{)}_2$ có các ion sau phản ứng: $N{H_4}^{+},S{O_4}^{2-},C{O_3}^{2-}$

$\begin{array}{c}N{H_4}^{+}+O{H}^{-}\to N{H}_3+H_{2}O\\ S{O_4}^{2-}+B{a}^{2+}\to BaS{O}_4\\ C{O_3}^{2-}+B{a}^{2+}\to BaC{O}_3\\ n_{N{H}_3}=0,02mol\\ C{O_3}^{2-}+2H^{+}\to C{O}_2+H_{2}O\\ n_{C{O}_2}=0,01mol\\ \to n_{C{O_3}^{2-}}=n_{C{O}_2}=0,01mol\\ \to m_{BaS{O}_4}=4,3-0,01\times 197=2,33g\\ \to n_{BaS{O}_4}=0,01mol=n_{S{O_4}^{2-}}\end{array}$

Bảo toàn điện tích trong dung dịch A, ta có:

$\begin{array}{c}{n}_{N{H}_3}=n_{N{H_4}^{+}}=0,02mol\\ n_{N{a}^{+}}+n_{N{H_4}^{+}}=2n_{S{O_4}^{2-}}+2n_{C{O_3}^{2-}}\\ \to n_{N{a}^{+}}=0,02mol\\ \to m=0,02\times 23+0,02\times 18+0,01\times 96+0,01\times 60\\ \to m=2,38g\end{array}$

- CuCl₂: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

- Cu(NO₃)₂: \(CuO+2HNO_3\rightarrow Cu\left(NO_3\right)_2+H_2O\)

- Fe₂(SO₄)₃: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

- FeCl₃: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

PT: \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

Ta có: \(n_{H_2SO_4}=0,25.1=0,25\left(mol\right)\)

a, Theo PT: \(n_{NaOH}=2n_{H_2SO_4}=0,5\left(mol\right)\Rightarrow V_{NaOH}=\dfrac{0,5}{2}=0,25\left(l\right)\)

b, Theo PT: \(n_{Na_2SO_4}=n_{H_2SO_4}=0,25\left(mol\right)\)

\(\Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,25}{0,25+0,25}=0,5\left(M\right)\)

Quá trình:

\(Fe_x^{^{+\dfrac{2y}{x}}}\rightarrow xFe^{+3}+\left(3x-2y\right)|\times4\)

\(N^{+5}+3e\rightarrow N^{+2}|\times\left(3x-2y\right)\) 

\(N^{+5}+1e\rightarrow N^{+4}|\times\left(3x-2y\right)\)

PT: \(4Fe_xO_y+\left(18x-4y\right)HNO_3\rightarrow4xFe\left(NO_3\right)_3+\left(3x-2y\right)NO+\left(3x-2y\right)NO_2+\left(9x-2y\right)H_2O\)