Bài 9: Axit nitric - Muối nitrat

\(n_{NO}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(3Mg+8HNO_3\rightarrow3Mg\left(NO_3\right)_2+2NO+4H_2O\)

 x          \(\dfrac{8}{3}x\)                  x                \(\dfrac{2}{3}x\)        \(\dfrac{4}{3}x\)

\(Fe+4HNO_3\rightarrow Fe\left(NO_3\right)_3+NO+2H_2O\)

 y         4y                   y                y      2y

dd Y: \(Mg\left(NO_3\right)_2,Fe\left(NO_3\right)_3\)

gọi x, y là số mol Mg, Fe

có hệ: \(\left\{{}\begin{matrix}24x+56y=9,2\\\dfrac{2}{3}x+y=0,2\end{matrix}\right.\)

<=> x = 0,15 và y = 0,1

a. \(V_{HNO_3}=\dfrac{\left(\dfrac{8}{3}.0,15+4.0,1\right).\left(100+20\right)}{2}:100=0,48\left(l\right)\)  

b. \(Mg+4HNO_{3.đn}\rightarrow Mg\left(NO_3\right)_2+2NO_2+2H_2O\)

    0,15                                                 0,3

  \(Fe+6HNO_{3.đn}\rightarrow Fe\left(NO_3\right)_3+3NO_2+3H_2O\)

 0,1                                                    0,3

\(V_{NO_2}=\left(0,3+0,3\right).22,4=13,44\left(l\right)\)

Cây 23:

a) Gọi \(n_{Mg\left(NO_3\right)_2\left(pư\right)}=a\left(mol\right);n_{khí}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: \(2Mg\left(NO_3\right)_2\xrightarrow[]{t^o}2MgO+4NO_2+O_2\)

             a---------------->a---------->2a---->0,5a

=> 2a + 0,5a = 0,6 => a = 0,24 (mol)

Mà \(n_{Mg\left(NO_3\right)_2\left(bđ\right)}=\dfrac{37}{148}=0,25\left(mol\right)\)

=> \(H=\dfrac{0,24}{0,25}.100\%=96\%\)

b) Chất rắn X có: \(\left\{{}\begin{matrix}n_{Mg\left(NO_3\right)_2\left(dư\right)}=0,25-0,24=0,01\left(mol\right)\\n_{MgO}=0,24\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{Mg\left(NO_3\right)_2}=0,24.40=9,6\left(g\right)\\m_{Mg\left(NO_3\right)_2\left(dư\right)}=0,01.148=1,48\left(g\right)\end{matrix}\right.\)

c) \(\overline{M}_Y=\dfrac{0,24.2.46+0,24.0,5.32}{0,6}=43,2\left(g/mol\right)\)

=> \(d_{Y/H_2}=\dfrac{43,2}{2}=21,6\)

d)                    \(4NO_2+O_2+2H_2O\rightarrow4HNO_3\)

Ban đầu:         0,48      0,12

Pư:                 0,48       0,12

Sau pư:         0               0                      0,48

=> \(C_{M\left(HNO_3\right)}=\dfrac{0,48}{2}=0,24M\)

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

a, Ta có \(n_{HNO_3p\text{/}ứ}=4n_{NO}=4.0,16=0,64\left(mol\right)=n_{H^+}\)

Ta có \(\dfrac{n_{HNO_3}}{n_{H_2SO_4}}=\dfrac{1}{1,5}=\dfrac{2V}{3V}\)

\(\Rightarrow2V+3V.2=0,64.22,4\) \(\Leftrightarrow V=1,792\left(l\right)\)

b, Ta có pt : 24x + 65y = 8,22

Bảo toàn e: 2x + 2y = 3.0,16 

\(\Rightarrow\left\{{}\begin{matrix}x=0,18\\y=0,06\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,18.24}{8,22}=52,55\%\\\%m_{Zn}=100\%-52,55\%=47,45\%\end{matrix}\right.\)

Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Cu}=b\left(mol\right)\end{matrix}\right.\) => 56a + 64b = 8,64 (*)

\(n_{khí.B}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)

Ta có: \(\overline{M_B}=19.2=38\left(g/mol\right)\)

=> \(m_B=0,16.38=6,08\left(g\right)\)

Gọi \(\left\{{}\begin{matrix}n_{NO}=x\left(mol\right)\\n_{NO_2}=y\left(mol\right)\end{matrix}\right.\) =>  \(\left\{{}\begin{matrix}x+y=0,16\\30x+46y=6,08\end{matrix}\right.\)

=> x = y = 0,08 (mol)

Quá trình oxi hóa:

\(Fe^0\rightarrow Fe^{+3}+3e\)

a--------------->3a

\(Cu^0\rightarrow Cu^{2+}+2e\)

b--------------->2b

=> \(\sum n_{e\left(nhường\right)}=3a+2b\)

Quá trình khử:

\(N^{+5}+1e\rightarrow N^{+4}\)

0,08->0,08

\(N^{+5}+3e\rightarrow N^{+2}\)

0,08->0,24

=> \(\sum n_{e\left(nhận\right)}=0,08+0,24=0,32\left(mol\right)\)

Bảo toàn electron: \(\sum n_{e\left(nhận\right)}=\sum n_{e\left(nhường\right)}\)

=> 3a + 2b = 0,32 (**)

Từ (*), (**) => a = 0,04; b = 0,1

=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,04.56}{8,64}.100\%=25,9\%\\\%m_{Cu}=100\%-25,9\%=74,1\%\end{matrix}\right.\)

b) BTNT Fe, Cu: \(\left\{{}\begin{matrix}n_{Fe\left(NO_3\right)_3}=n_{Fe}=0,04\left(mol\right)\\n_{Cu\left(NO_3\right)_2}=n_{Cu}=0,1\left(mol\right)\end{matrix}\right.\)

=> m_muối = 0,04.242 + 0,1.188 = 28,48 (g)

BTNT N: \(n_{HNO_3}=3n_{Fe\left(NO_3\right)_3}+2n_{Cu\left(NO_3\right)_2}+n_{NO}+n_{NO_2}=0,48\left(mol\right)\)

=> \(V_{ddHNO_3}=\dfrac{0,48}{2}=0,24\left(l\right)=240\left(ml\right)\)

\(n_{FeO}=\dfrac{21,6}{72}=0,3\left(mol\right);n_X=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Gọi số oxi hóa của N trong X là n

BTe: \(n_{FeO}=\left(5-n\right).n_X\)

=> \(5-n=\dfrac{0,3}{0,1}=3\)

=> n = 2

=> X là NO

Câu 29:

a, Ta có: \(n_{NO}=0,02\left(mol\right)\)

Theo ĐLBT e, có: \(3n_{Fe}=3n_{NO}\Rightarrow n_{Fe}=0,02\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,02.56}{1,52}.100\%\approx73,68\%\\\%m_{MgO}\approx26,32\%\end{matrix}\right.\)

b, Ta có: \(n_{MgO}=\dfrac{1,52-0,02.56}{40}=0,01\left(mol\right)\)

BTNT Fe: \(n_{Fe\left(NO_3\right)_3}=n_{Fe}=0,02\left(mol\right)\)

BTNT Mg: \(n_{Mg\left(NO_3\right)_2}=n_{MgO}=0,01\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{Fe\left(NO_3\right)_3}}=\dfrac{0,02}{0,2}=0,1\left(M\right)\\C_{M_{Mg\left(NO_3\right)_2}}=\dfrac{0,01}{0,2}=0,05\left(M\right)\end{matrix}\right.\)

BTNT N, có: \(n_{HNO_3\left(pư\right)}=3n_{Fe\left(NO_3\right)_3}+2n_{Mg\left(NO_3\right)_2}+n_{NO}=0,1\left(mol\right)\)

Mà: \(n_{HNO_3\left(banđau\right)}=0,2.1=0,2\left(mol\right)\)

\(\Rightarrow n_{HNO_3\left(dư\right)}=0,1\left(mol\right)\Rightarrow\left[H^+\right]=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

\(\Rightarrow pH=-\log\left(0,5\right)\approx0,3\)

c, \(H_{\left(dư\right)}^++OH^-\rightarrow H_2O\)

_0,1______0,1 (mol)

\(Fe^{3+}+3OH^-\rightarrow Fe\left(OH\right)_3\)

0,02_____0,06 (mol)

\(Mg^{2+}+2OH^-\rightarrow Mg\left(OH\right)_2\)

0,01_____0,02 (mol)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,1+0,06+0,02}{0,2}=0,9\left(M\right)\)

a, Ta có: \(n_{NO_2}=2,1\left(mol\right)\)

Gọi: \(\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Zn}=y\left(mol\right)\end{matrix}\right.\)

⇒ 56x + 65y = 47,5 (1)

Theo ĐLBT e, có: 3x + 2y = 2,1 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,5\left(mol\right)\\y=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,5.56}{47,5}.100\%\approx58,95\%\\\%m_{Zn}\approx41,05\%\end{matrix}\right.\)

b, Ta có: m _{dd sau pư} = 47,5 + 441 - 2,1.46 = 391,9 (g)

BTNT Fe: n_Fe(NO3)3 = n_Fe = 0,5 (mol)

BTNT Zn: n_Zn(NO3)2 = n_Zn = 0,3 (mol)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Fe\left(NO_3\right)_3}=\dfrac{0,5.242}{391,9}.100\%\approx30,88\%\\C\%_{Zn\left(NO_3\right)_2}=\dfrac{0,3.189}{391,9}.100\%\approx14,47\%\end{matrix}\right.\)

Ta có: \(n_{NO_2}=0,4\left(mol\right)\)

a, PT: \(Cu+4HNO_3\rightarrow Cu\left(NO_3\right)_2+2NO_2+2H_2O\)

_____0,2_____0,8_________________0,4 (mol)

b, \(m=m_{Cu}=0,2.64=12,8\left(g\right)\)

\(a=C_{M_{HNO_3}}=\dfrac{0,8}{0,2}=4\left(M\right)\)