Bài 23: Hiđro Clorua - Axit Clohidric và muối Clorua

ta có 200cm3=0,2 lítn hcl=2*0,2=0,4 molgọi số mol của caco3 là a,na2co3 là bcaco3 + 2hcl -> cacl2 + co2 + h2oa(mol)---2a(mol)--a-------a--------ana2co3 + 2hcl -> 2nacl + co2 + h2ob(mol)---2b(mol)---2b-------b------bta có100a+106b=20,62a+2b=0,4=> a=b=0,1 mol=> m caco3=10g; m na2co3=10,6 g 

PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\)  (1)

            \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)  (2)

a) Ta có: \(\Sigma n_{HCl}=0,2\cdot2=0,4\left(mol\right)\)

Gọi số mol của Na₂CO₃ là \(a\) \(\Rightarrow n_{HCl\left(1\right)}=2a\left(mol\right)\)

Gọi số mol của CaCO₃ là \(b\) \(\Rightarrow n_{HCl\left(2\right)}=2b\left(mol\right)\)

Ta lập được hệ phương trình:

\(\left\{{}\begin{matrix}2a+2b=0,4\\106a+100b=20,6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=0,1\cdot100=10\left(g\right)\\m_{Na_2CO_3}=10,6\left(g\right)\end{matrix}\right.\)

b) Theo PTHH: \(\left\{{}\begin{matrix}n_{NaCl}=2n_{Na_2CO_3}=0,2mol\\n_{CaCl_2}=n_{CaCO_3}=0,1mol\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,2\cdot58,5=11,7\left(g\right)\\m_{CaCl_2}=0,1\cdot111=11,1\left(g\right)\end{matrix}\right.\)

Mặt khác: \(\left\{{}\begin{matrix}m_{CO_2}=0,2\cdot44=8,8\left(g\right)\\m_{ddHCl}=200\cdot1,2=240\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{dd\left(saup/ư\right)}=m_{hh}+m_{ddHCl}-m_{CO_2}=251,8\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{11,7}{251,8}\cdot100\%\approx4,65\%\\C\%_{CaCl_2}=\dfrac{11,1}{251,8}\cdot100\%\approx4,41\%\end{matrix}\right.\)

PTHH: \(H_2+Cl_2\underrightarrow{as}2HCl\)

Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\n_{Cl_2}=\dfrac{0,71}{71}=0,01\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) H₂ còn dư, Cl₂ p/ứ hết

\(\Rightarrow n_{HCl\left(lýthuyết\right)}=0,02mol\) \(\Rightarrow n_{HCl\left(thựctế\right)}=0,008mol\)

\(\Rightarrow V_{HCl}=0,008\cdot22,4=0,1792\left(l\right)\) 

nH2= 0,03/2= 0,015(mol)

nCl2=0,71/71= 0,01(mol)

PTHH: H2 + Cl2 -to-> 2 HCl

Ta có: 0,015/1 > 0,01/1

=> H2 dư, Cl2 hết, tính theo nCl2.

Vì H=40% nên ta có:

nHCl= 2. nCl2. H= 2. 0,01. 40%= 0,008(mol)

=>V(HCl,đktc)= 0,008.22,4=0,1792(l)

=> CHỌN C

Gọi : \(\left\{{}\begin{matrix}n_{Al_2O_3}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\)⇒ 102a + 65b = 2,505(1)

\(Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O\\ Zn + 2HCl \to ZnCl_2 + H_2\)

Muối gồm : \(\left\{{}\begin{matrix}AlCl_3:2a\left(mol\right)\\ZnCl_2:b\left(mol\right)\end{matrix}\right.\)⇒ 133,5.2a + 136b = 6,045(2)

Từ (1)(2) suy ra : a = 0,015 ; b = 0,015

Vậy :

\(\%m_{Al_2O_3} = \dfrac{0,015.102}{2,505}.100\% = 61,08\%\\ \%m_{Zn} = 100\% - 61,08\% = 38,92\%\)

Theo PTHH : \(n_{HCl} = 6a + 2b = 0,12(mol)\\ \Rightarrow C\%_{HCl} = \dfrac{0,12.36,5}{200}.100\% = 2,19\%\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)  (1)

            \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)  (2)

Ta có: \(\Sigma n_{HCl}=1\cdot0,5=0,5\left(mol\right)\)

Gọi số mol của Fe là \(a\) \(\Rightarrow n_{HCl\left(1\right)}=2a\)

Gọi số mol của Fe₂O₃ là b \(\Rightarrow n_{HCl\left(2\right)}=6b\)

Ta lập được hệ phương trình:

\(\left\{{}\begin{matrix}2a+6b=0,5\\56a+160b=13,6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{FeCl_2}=0,1mol\\n_{FeCl_3}=0,1mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\\m_{FeCl_3}=0,1\cdot162,5=16,25\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{muối}=12,7+16,25=28,95\left(g\right)\)

Đặt : 

nFe = x mol 

nFe2O3 = y mol 

mhh = 56x + 160y = 13.6 (g) (1) 

Fe + 2HCl => FeCl2 + H2 

x_____2x

Fe2O3 + 6HCl => 2FeCl3 + 3H2O 

y________6y

nHCl = 2x + 6y = 0.5 (2) 

(1) , (2) : 

x = 0.1

y = 0.05 

mFeCl2 = 0.1*127 = 12.7 (g) 

mFeCl3 = 0.1*162.5 = 16.25 (g) 

a)\(\left\{{}\begin{matrix}Fe:a\left(mol\right)\\Al:b\left(mol\right)\end{matrix}\right.\)⇒ 56a + 27b = 1,93(1)

\(Fe + 2HCl \to FeCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\)

Theo PTHH : a + 1,5b = \(\dfrac{1,456}{22,4} = 0,065\)(2)

Từ (1)(2) suy ra : a = 0,02 ; b = 0,03

Vậy  :

\(\%m_{Fe} = \dfrac{0,02.56}{1,93}.100\% = 58,03\%\\ \%m_{Al} = 100\% - 58,03\% = 41,97\%\)

b) 

\(C_{M_{FeCl_2}} = \dfrac{0,02}{0,2} = 0,1M\\ C_{M_{AlCl_3}} = \dfrac{0,03}{0,2} = 0,15M\)

c)

\(n_{HCl} = 2n_{H_2} = 0,065.2 = 0,13(mol)\\ a = \dfrac{0,13}{0,2} = 0,65(M)\)

mn thêm vào đề bài giúp mình là hỗn hợp gồm 1,93g Fe,Al nha

\(n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\)

\(\left\{{}\begin{matrix}Mg:x\left(mol\right)\\Al:y\left(mol\right)\end{matrix}\right.\)→ \(\left\{{}\begin{matrix}24x+27y=6,3\\x+1,5y=0,3\end{matrix}\right.\)→ \(\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)

Vậy :

\(\%m_{Mg} = \dfrac{0,15.24}{6,3}.100\% = 57,14\%\)

a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Ta có: \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

Theo PT: \(n_{Mg}=n_{H_2}=0,03\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,03.24=0,72\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,72}{1,74}.100\%\approx41,38\%\\\%m_{AlCl_3}\approx58,62\%\end{matrix}\right.\)

b, Theo PT: \(n_{HCl}=2n_{H_2}=0,06\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,06.36,5=2,19\left(g\right)\)

\(\Rightarrow C\%_{ddHCl}=\dfrac{2,19}{500}.100\%=0,438\%\)

Bạn tham khảo nhé!

a, nH2 = 0,03 ( mol )

=> nMg = nH2 = 0,03 ( mol )

=> mMg = 0,72 g

=> %Mg \(\approx\) 41,38 % .

=> % Al \(\approx\) 58,62 % .

b, Có : nH2 = 0,03 mol

=> nHCl = nHCl_{từ Al2O3} + nHCl_{từ Mg} = 0,06 + 0,06 = 0,12 ( mol )

=> mHCl = 4,38 ( g )

Lại có : mdd = mhh + mddHCl = 501,74 ( g )

=> \(C\%=\dfrac{m_{HCl}}{m_{dd}}.100\%\approx0,87\%\)

( chắc đoạn trên là Al2O3 :vvvv )

Bài 33:

a, Ta có Ag không tác dụng với dd HCl.

⇒ m_cr = m_Ag = 4,6 (g)

⇒ m_Al = 10 - 4,6 = 5,4 (g)

b, \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{5,4}{10}.100\%=54\%\\\%m_{Ag}=46\%\end{matrix}\right.\)

c, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

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

\(\Rightarrow V=V_{H_2}=0,3.22,4=6,72\left(l\right)\)

d1, Theo PT: \(n_{HCl}=3n_{Al}=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

d2, Theo PT: \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

e, Ta có: \(m_{ddHCl}=\dfrac{21,9.100}{10}=219\left(g\right)\)

⇒ m _{dd sau pư} = m_Al + m _ddHCl - m_H2 = 5,4 + 219 - 0,3.2 = 223,8 (g)

\(\Rightarrow C\%_{AlCl_3}=\dfrac{26,7}{223,8}.100\%\approx11,93\%\)

Bạn tham khảo nhé!

giúp em câu 33

n_HCl = 0.4 (mol) 

n_Ba(OH)2 = 0.15 (mol) 

Ba(OH)₂ + 2HCl => BaCl₂ + H₂O

0.15_______0.3_____0.15 

C_{M HCl(dư)} = 0.3/0.4 = 0.75M

C_{M BaCl2} = 0.15/(0.4 + 0.3) = 3/14 (M) 

\(Đặt:n_{MnO_2}=a\left(mol\right),n_{KMnO_4}=b\left(mol\right)\)

\(m_{hh}=87a+158b=37.96\left(g\right)\left(1\right)\)

\(n_{Cl_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)

\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)

\(n_{Cl_2}=a+2.5b=0.45\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.4,b=0.02\)

\(\%MnO_2=\dfrac{0.4\cdot87}{37.96}\cdot100\%=91.68\%\\\%KMnO_4=100-91.68=8.32\% \)

\(m_M=m_{KCl}+m_{MnCl_2}=0.02\cdot74.5+\left(0.4+0.02\right)\cdot126=54.41g\)