Bài 14: Bài thực hành 3

a) Fe + 2HCl ---> FeCl₂ + H₂

b) Tỉ lệ số nguyên tử Fe : số phân tử HCl : số phân tử FeCl₂ : số phân tử H₂ = 1 : 2 : 1 : 1

muói sắt (II) clorua phải là FeCl2 chứ nhỉ?

Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\) => 27a + 56b = 21,9 (*)

\(n_{O_2}=\dfrac{10,64}{22,4}=0,475\left(mol\right)\)

PTHH:

\(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)

a----->0,75a--->0,5a

\(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)

b----->\(\dfrac{2}{3}b\)---->\(\dfrac{1}{3}b\)

=> \(0,75a+\dfrac{2}{3}b=0,475\) (**)

Từ (*), (**) => \(\left\{{}\begin{matrix}a=0,5\\b=0,15\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,5.27}{21,9}.100\%=61,64\%\\\%m_{Fe}=100\%-61,64\%=38,36\%\end{matrix}\right.\)

m_{chất rắn sau phản ứng} = 21,9 + 0,475.32 = 37,1 (g)

=> \(\left\{{}\begin{matrix}\%m_{Al_2O_3}=\dfrac{0,5.102.0,5}{37,1}.100\%=68,73\%\\\%m_{Fe_3O_4}=100\%-68,73\%=31,27\%\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Fe}=y\end{matrix}\right.\) ( mol )

\(\rightarrow m_{hh}=27x+56y=21,9\left(g\right)\) (1)

\(n_{O_2}=\dfrac{10,64}{22,4}=0,475\left(mol\right)\)

\(4Al+3O_2\rightarrow\left(t^o\right)Al_2O_3\)

 `x`     `3/4 x`                  `1/4 x`      ( mol )

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)

 `y`     `2/3 y`                  `1/3 y`       ( mol )

\(\rightarrow n_{O_2}=\dfrac{3}{4}x+\dfrac{2}{3}y=0,475\left(mol\right)\) (2)

\(\left(1\right);\left(2\right)\rightarrow\left\{{}\begin{matrix}x=0,5\\y=0,15\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,5.27}{21,9}.100=61,64\%\\\%m_{Fe}=100\%-61,64\%=38,36\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}m_{Al_2O_3}=\dfrac{1}{4}.0,5.102=12,75\left(g\right)\\m_{Fe_3O_4}=\dfrac{1}{3}.0,15.232=11,6\left(g\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{Al_2O_3}=\dfrac{12,75}{12,75+11,6}.100=52,36\%\\\%m_{Fe_3O_4}=100\%-52,36\%=47,64\%\end{matrix}\right.\)

Bài 3:

Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\) => 27a + 56b = 21,9 (*)

\(n_{O_2}=\dfrac{10,64}{22,4}=0,475\left(mol\right)\)

PTHH:
\(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)

a----->0,75a-->0,5a

\(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)

b---->\(\dfrac{2}{3}b\)------->\(\dfrac{1}{3}b\)

=> \(0,75a+\dfrac{2}{3}b=0,475\) (**)

Từ (*), (**) => a = 0,5; b = 0,15

=> Trước phản ứng \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,5.27}{21,9}.100\%=61,64\%\\\%m_{Fe}=100\%-61,64\%=38,36\%\end{matrix}\right.\)

m_{chất rắn sau phản ứng} = 21,9 + 0,475.32 = 37,1 (g)

=> Sau phản ứng \(\left\{{}\begin{matrix}\%m_{Al_2O_3}=\dfrac{0,5.0,5.102}{37,1}.100\%=68,73\%\\\%m_{Fe_3O_4}=100\%-68,73\%=31,27\%\end{matrix}\right.\)

`a)`

Gọi \(n_K=a\left(mol\right)\)

PTHH:

                   \(2K+2H_2O\rightarrow2KOH+H_2\uparrow\)

                   a----------------->a

              \(2KOH+2Al+2H_2O\rightarrow2KAlO_2+3H_2\uparrow\)

                a--------->a

\(\Rightarrow\dfrac{n_K}{n_{Al}}=\dfrac{a}{a}=\dfrac{1}{1}\)

`b)` 

Gọi \(\left\{{}\begin{matrix}n_K=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow39a+27b=14,4\left(1\right)\)

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

PTHH:

                      \(2K+2HCl\rightarrow2KCl+H_2\uparrow\)

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

                      \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

                      b------------------------------>1,5b

\(\Rightarrow0,5a+1,5b=0,3\left(2\right)\\ \left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=0,3\left(mol\right)\\b=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_K=0,3.39=11,7\left(g\right)\\m_{Al}=0,1.27=2,7\left(g\right)\end{matrix}\right.\)

`c)`

\(n_{Al\left(OH\right)_3}=\dfrac{1,56}{78}=0,02\left(mol\right)\)

\(\xrightarrow[]{\text{BTNT Al}}n_{NaAlO_2}=0,1-0,02=0,08\left(mol\right)\)

\(n_{HCl\left(bđ\right)}=2.0,4=0,8\left(mol\right)\)

PTHH:

                      \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

                      0,1---->0,3------>0,1

                      \(2K+2HCl\rightarrow2KCl+H_2\)

                      0,3-->0,3-------->0,3

 \(\Rightarrow n_{HCl\left(dư\right)}=0,8-0,3-0,3=0,2\left(mol\right)\)

PTHH: 

                       \(NaOH+HCl\rightarrow NaCl+H_2O\)

                        0,2<------0,2

                       \(3NaOH+AlCl_3\rightarrow Al\left(OH\right)_3\downarrow+3NaCl\)

                        0,3<-------0,1

                       \(NaOH+Al\left(OH\right)_3\rightarrow NaAlO_2+2H_2O\)

                        0,08<--------------------0,08

\(\Rightarrow C_{M\left(NaOH\right)}=\dfrac{0,3+0,2+0,08}{2}=0,29M\)

hỏi 1 lần thôi bn :)

a)

Gọi số mol K, Al là a, b (mol)

PTHH: 2K + 2H₂O --> 2KOH + H₂

            a---------------->a

            2Al + 2KOH + 2H₂O --> 2KAlO₂ + 3H₂

              b---->b

Để hỗn hợp tan hết \(\Leftrightarrow a\ge b\Leftrightarrow\dfrac{a}{b}\ge1\)

b) 

Ta có: 39a + 27b = 14,4 (1)

PTHH: 2K + 2H₂O --> 2KOH + H₂

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

            2Al + 2KOH + 2H₂O --> 2KAlO₂ + 3H₂

              b---->b--------------------------->1,5b

=> 0,5a + 1,5b = \(\dfrac{6,72}{22,4}=0,3\left(mol\right)\) (2)

(1)(2) => a = 0,3 (mol); b = 0,1 (mol)

\(\left\{{}\begin{matrix}m_K=0,3.39=11,7\left(g\right)\\m_{Al}=0,1.27=2,7\left(g\right)\end{matrix}\right.\)

c)

n_HCl = 2.0,4 = 0,8 (mol)

PTHH: 2K + 2HCl --> 2KCl + H₂

           0,3-->0,3

           2Al + 6HCl --> 2AlCl₃ + 3H₂

            0,1-->0,3----->0,1

=> X chứa \(\left\{{}\begin{matrix}KCl\\AlCl_3:0,1\left(mol\right)\\HCl\left(dư\right):0,2\left(mol\right)\end{matrix}\right.\)

Gọi số mol NaOH phản ứng là x (mol)

\(n_{Al\left(OH\right)_3}=\dfrac{1,56}{78}=0,02\left(mol\right)\)

TH1: Kết tủa không bị hòa tan

PTHH: NaOH + HCl --> NaCl + H₂O

              0,2<---0,2

            3NaOH + AlCl₃ --> 3NaCl + Al(OH)₃

              0,06<------------------------0,02

=> x = 0,2 + 0,06 = 0,26 (mol)

=> \(C_{M\left(dd.NaOH\right)}=\dfrac{0,26}{2}=0,13M\)

TH2: Kết tủa bị hòa tan 1 phần 

PTHH: NaOH + HCl --> NaCl + H₂O

              0,2<---0,2

            3NaOH + AlCl₃ --> 3NaCl + Al(OH)₃

                0,3<----0,1----------------->0,1

            Al(OH)₃ + NaOH --> NaAlO₂ + 2H₂O

              0,08---->0,08

=> x = 0,2 + 0,3 + 0,08 = 0,58 (mol)

=> \(C_{M\left(dd.NaOH\right)}=\dfrac{0,58}{2}=0,29M\)

Gọi số mol CO, O₂, N₂ là a, b, 4b (mol)

\(\overline{M}=\dfrac{28a+32b+28.4b}{a+b+4b}=7,12.4=28,48\left(g/mol\right)\)

=> \(28a+144b=28,48a+142,4b\)

=> \(0,48a=1,6b\)

=> \(b=0,3a\)

\(\%V_{CO}=\dfrac{a}{a+b+4b}.100\%=\dfrac{a}{a+0,3a+1,2a}.100\%=40\%\)

\(\%V_{O_2}=\dfrac{b}{a+b+4b}.100\%=\dfrac{0,3a}{a+0,3a+1,2a}.100\%=12\%\)

\(\%V_{N_2}=\dfrac{4b}{a+b+4b}.100\%=\dfrac{1,2a}{a+0,3a+1,2a}.100\%=48\%\)

\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(R+2HCl\rightarrow RCl_2+H_2\)

\(0.1...............................0.1\)

\(M_R=\dfrac{5.6}{0.1}=56\left(\dfrac{g}{mol}\right)\)

\(R:Fe\left(Sắt\right)\)

\(a,PTHH:CaCO_3\xrightarrow{t^o}CaO+CO_2\\ b,BTKL:m_{CaCO_3}=m_{CaO}+m_{CO_2}\\ c,m_{CaO}=m_{CaCO_3}-m_{CO_2}=10-4,4=5,6(tấn)\)

á à thi dg k

\(n_{CO_2}=\dfrac{PV}{RT}=\dfrac{0.99\cdot2.479}{0.082\cdot\left(273+25\right)}=0.1\left(mol\right)\)

\(1bar=0.99atm\)

\(a.Nhôm+Axitclohidric\rightarrow Nhômclorua+Hidro\\ b.2Al+6HCl\rightarrow2AlCl_3+3H_2\\ c.m_{Al}+m_{HCl}=m_{AlCl_3}+m_{H_2}\\ \Rightarrow m_{H_2}=13,5+73-66,75=19,75\left(g\right)\)