Chương 2. Nhóm nitơ

\(n_{Cu}=\dfrac{9,6}{64}=0,15\left(mol\right)\)

\(n_{HNO_3}=0,18\left(mol\right)\)

\(+TH_1:\)

Vì thu được \(NO\) duy nhất → Có 2 trường hợp xảy ra, hoặc là `Cu` hết hoặc là `HNO_3` hết

\(\left[e\right]:2.0,15=3.n_{NO}\Leftrightarrow n_{NO}=0,1\left(mol\right)\)

Mặt khác ta có: \(n_{H^+}=4n_{NO}=0,18\Leftrightarrow n_{NO}=0,045\left(mol\right)\)

→ Chứng tỏ `Cu` dư 

\(NO\left(0,045\right)\underrightarrow{+O_2}NO_2\rightarrow HNO_3\)

\(PTHH:NO_2+\dfrac{1}{2}O_2+H_2O\rightarrow2HNO_3\)

\(\left(mol\right)\)   \(0,045\)       \(0,0225\)

→ Tính theo oxi
\(\Rightarrow n_{HNO_3}=0,0225.2.2=0,09\left(mol\right)\\ \Rightarrow\left[H^+\right]=\dfrac{0,09}{0,2}=0,45\left(M\right)\\ \Rightarrow pH=-log\left(H^+\right)\approx0,35\)

\(+TH_2:\)  Tóm tắt các phản ứng như sau

\(0,15\left(mol\right)Cu+\left\{{}\begin{matrix}H_2SO_4:0,09\left(mol\right)\\HNO_3:0,18\left(mol\right)\end{matrix}\right.\rightarrow V\left(l\right)NO+B\left\{{}\begin{matrix}Cu\left(NO_3\right)_2:0,135\\SO_4^{2-}:0,09\end{matrix}\right.+Cu:0,015\left(mol\right)\)

Tương tự như trường hợp 1 ta có:

\(n_{H^+}=4n_{NO}\Leftrightarrow0,36=4n_{NO}\Leftrightarrow n_{NO}=0,09\left(mol\right)\)  và \(Cu\) dư \(0,015\left(mol\right)\)

 \(\left\{{}\begin{matrix}0,045\left(mol\right)Cu\left(NO_3\right)_2\\0,09\left(mol\right)CuSO_4\end{matrix}\right.+Cu\left(0,015\right)\underrightarrow{t^o}\left\{{}\begin{matrix}CuO:0,045\\CuSO_4:0,09\\Cu:0,015\end{matrix}\right.\)

\(\rightarrow m=0,45.80+0,09.160+0,015.64=18,96\left(g\right)\)

\(n_{NO}=a\left(mol\right),n_{N_2O}=b\left(mol\right)\)

\(n_{hh}=a+b=\dfrac{8.96}{22.4}=0.4\left(mol\right)\left(1\right)\)

\(M_A=20.2=40\left(\dfrac{g}{mol}\right)\)

\(m_A=30a+44b=40\cdot0.4=16\left(g\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=\dfrac{4}{35},b=\dfrac{2}{7}\)

Bảo toàn e : 

\(n_{Al}=\dfrac{\dfrac{4}{35}\cdot3+\dfrac{2}{7}\cdot8}{3}=0.87\left(mol\right)\)

\(m_{Al}=23.49\left(g\right)\)

a, \(\left\{{}\begin{matrix}27n_{Al}+64n_{Cu}=2,46\\3n_{Al}+2n_{Cu}=3n_{NO}=0,12\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,02\left(mol\right)\\n_{Cu}=0,03\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,54\left(g\right)\\n_{Cu}=1,92\left(g\right)\end{matrix}\right.\)

b, \(n_{HNO_3}=4n_{NO}=0,16\left(mol\right)\)

\(\Rightarrow m_{HNO_3}=10,08\left(g\right)\)

\(\Rightarrow m_{ddHNO_3}=201,6\left(g\right)\)

\(\Rightarrow m_{\text{dd sau pư}}=2,46+201,6-0,04.30=202,86\left(g\right)\)

\(\Rightarrow C\%\left(Al\left(NO_3\right)_3\right)=\dfrac{0,02.213}{202,86}.100\%=2,1\%\)

\(\Rightarrow C\%\left(Cu\left(NO_3\right)_2\right)=\dfrac{0,03.188}{202,86}.100\%=2,78\%\)

\(\left\{{}\begin{matrix}n_{N_2O}+n_{NO}=0,04\\44n_{N_2O}+30n_{NO}=18,5.2.0,04=1,48\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{N_2O}=0,02\left(mol\right)\\n_{NO}=0,02\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{N_2O}=\dfrac{0,02.44}{1,48}.100\%=59,46\%\\\%m_{NO}=\dfrac{0,02.30}{1,48}.100\%=40,54\%\end{matrix}\right.\)

\(Zn\rightarrow Zn^{2+}+2e\\ 4H^++NO_3^-+3e\rightarrow NO+2H_2O\\ CuO+HNO_3\rightarrow Cu\left(NO_3\right)_2+H_2O\\ n_{NO}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\\ BTe:n_{Zn}.2=n_{NO}.3\\ \Rightarrow n_{Zn}=0,03\left(mol\right)\\ \Rightarrow m_{CuO}=2,75-0,03.65=0,8\left(g\right)\\ \Rightarrow n_{CuO}=\dfrac{0,8}{80}=0,01\left(mol\right)\\ n_{HNO_3}=n_{NO}.4+n_{CuO}.2=0,1\left(mol\right)\\ \Rightarrow CM_{HNO_3}=\dfrac{0,1}{0,2}=0,5M\\ \Rightarrow ChọnA\)

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hh X gồm Fe 7,28 g và O x mol

Tính đc số mol:

+n_Fe= 0,13 mol

+n_khí = 0,07 mol

Vì axit là Đn nên Khí là NO2

BTe ta có : 3.n_Fe -2.n_O = n_NO2---> n_O=0,16 mol

=> m = m_Fe + m_O= 9,84 g

\(n_{Mg}=\dfrac{38,4}{24}=1,6\left(mol\right);n_Z-\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ Mg\rightarrow Mg^{2+}+2e\\ N^{+5}+ne\rightarrow N_xO_y\\ Bảotoàne:1,6.2=0,4.n\\ \Rightarrow n=8\\ \Rightarrow KhíZlàN_2O\)